Vectors - solutions.pdf

RRHS PHYSICS 12TABLE OF CONTENTSModule 1.1 Displacement and Velocity Vectors ....................................................................... 31.1.1 Introduction to 2D Vectors ...................................................................................... 31.1.2 Displacement Vectors and Vector Algebra .......................................................... 131.1.3 Velocity Vectors .................................................................................................... 331.1.4 Module Summary ................................................................................................. 53Module 1.2 Force Vectors.......................................................................................................... 541.2.1 Dynamics Problems in 2D .................................................................................... 541.2.2 Inclined Planes ..................................................................................................... 691.2.3 Module Summary ................................................................................................. 89Module 1.3 Equilibrium .............................................................................................................. 901.3.1 Translational Equilibrium ...................................................................................... 901.3.2 Torque and Rotational Equilibrium ..................................................................... 1061.3.3 Module Summary ............................................................................................... 128Module 1.4 2D Collisions ......................................................................................................... 1291.4.1 Conservation of Momentum ............................................................................... 1291.4.2 Elastic and Inelastic Collisions ........................................................................... 141RRHS Physics Page 2

Module 1.1 Displacement and Velocity Vectors1.1.1 Introduction to 2D VectorsIn unit 2, you discussed two kinds of quantities --- vectors and scalars. A scalar is aquantity that has only magnitude (size); it does not have a direction. For example,temperature and mass have no direction associated with them. A vector is a quantitythat has both magnitude and direction. For example, displacement, velocity,acceleration, force, and momentum are all quantities for which it is important to knowthe direction. A vector quantity is denoted by placing an arrow over the symbol ( d );when typing, a vector quantity can be denoted by using boldface (d). In this manual,we will use the arrow notation.In units 2 and 3, you talked briefly about vectors in one dimension. In one dimension,we assigned directions to vectors using positive and negative signs. In this unit, wewill be extending that analysis to two dimensions. A vector in two dimensions is notjust a single number. In two-dimensional space a vector is actually represented withtwo numbers, one for each dimension. You have used an x y coordinate system inmath, and you know that two numbers are needed to specify a position on one ofthese graphs. Likewise, two coordinates are needed to specify a vector in twodimensionalspace. These coordinates are referred to as components. Consider thediagram below, where east is represented by the positive x direction and north isrepresented by the positive y direction. 11 In physics we usually use compass directions in conjunction with a traditional x-y plane since we aredealing with the actual motion of objects in real space.RRHS Physics Page 3

dFigure 1: Vector ComponentsThe position vector d in Figure 1 represents a step in space from the origin to somepoint whose location is given by the ordered pair ( dx, dy)- dxand dyare known asthe components of d . As with one dimensional vectors, the direction of componentscan be specified with a positive or negative sign. The symbol d represents both ofthese components together. We will use the symbols dxand dyto refer to themagnitude of the components; the symbolsdirection by using a positive or negative sign.dxanddywill be used to include theInstead of representing a two-dimensional vector by using both of its components, itis often convenient to represent a vector by an arrow that indicates the direction ofthe vector. The vector d shown in Figure 1 can then be described using a magnituded (the ``length'' of the vector) and an angle (the direction of the vector). Note thatif we know the magnitude d and the angle , we can use sin and cos functionsto solve for dxand dyin the above diagram.Trigonometry ReviewConsider the following right angle triangle:RRHS Physics Page 4

Figure 2: Right Angle Triangle Trigonometry ReviewWhere The hypotenuse is the side across from the right angle (the longest side). The opposite side is the side across from the angle θ. The adjacent side is the side joining the angle θ and the right angle.adjacentoppositeoppositecos sin tan hypotenusehypotenuseadjacentAlso, remember that the lengths of the sides can be related to one another using thePythagorean Theorem2 2 2c a b or2 2c a bwhere c is the hypotenuse of the triangle.ExampleIn Figure 1, assume that the angle in the diagram is 30.0 and the magnitude of theposition vector is 7.2 cm. Calculate the components dxand dy.SolutionRRHS Physics Page 5

By definition, components must be perpendicular to one another 2 ; therefore, theabove diagram is a right angle triangle and we can apply the basic trigonometricfunctions. Since we know the hypotenuse of the triangle (7.2 cm), we can use thecosine function to calculate the adjacent side of the triangle ( dx) and the sinefunction to calculate the opposite side of the triangle ( d ):adjcoshypdxcosddxcos30.0 7.2dx 6.2cmsindysinddysin 30.0 7.2dyopphypy 3.6cmIn other words, the location identified by the position vector identified by d can alsobe reached by going 6.2 cm to the east and 3.6 cm to the north. When calculatingcomponents, we will often need to identify the direction of the components usingpositive or negative signs. East (positive x-direction) and north (positive y-direction)will be considered positive. West (negative x-direction) and south (negative y-direction) will be considered negative. For the example above, since the directions ofthe components are east and north, both should be positive.In the above example, we drew a sketch of the vector and used trigonometry tocalculate the components. Vectors can also be drawn using scale diagrams, where aprotractor can be used to orient the vector correctly and an appropriate scale can beused to represent the vector. For example, a scale of 1 cm for every 5 m can beused; a 30 m displacement vector would then be drawn with an arrow that is 6 cmlong. The components can then be drawn in and measured with a ruler.When drawing a vector, the arrow indicates the direction of the vector. The arrow isreferred to as the head of the vector and the tail is at the other end.2 Since they are parallel to the axes in the coordinate system.RRHS Physics Page 6

Figure 3: Vector head and tailRRHS Physics Page 7

Check Your LearningCalculate the components for each of the following vectors:a)b)dxcos5512.2d 7.0mxdysin 5512.2d 10.myTaking the signs into consideration (to the right and down), we getdx 7.0mand d 10. mydxsin 438.5d 5.8cmxdycos 438.5d 6.2cmTaking the signs into consideration (to the left and up), we getdx 5.8cmand d 6.2cmyyRRHS Physics Page 8

Expressing DirectionThere are different conventions for describing the direction of a vector. For theexamples that follow, assume that 30 o in Figure 1.1. In math, you may have described vector directions as a counterclockwiserotation from the positive x-coordinate (east using compass directions). InFigure 1, the direction of the vector would then be 30 o . Using this convention,north would be 90 o and south would be 270 o .2. Bearings are another way of expressing directions and are often used innavigation. In this system, north is 0 o and all directions are measuredclockwise from this reference direction. In this system, the direction of thevector in our diagram would be 60 o .3. The last convention we will discuss is the one that we are going to use. Thisconvention describes a direction as a rotation from one of the four referencedirections (north, east, south, west). The direction of the vector in our diagramwould now be 30 o north of east. (or 30 o N of E) This means that a vector thatwas pointed east was rotated 30 o toward the north. This convention isconvenient because there is no ambiguity about what the reference direction (0 o ) is. The direction in the diagram could also be expressed as 60 o east ofnorth. A slightly different way of expressing 30 o north of east would be to sayE30 o N - this can be interpreted as “go east and then rotate 30 o toward thenorth” for the proper vector direction. The textbook that accompanies thismanual uses this last convention.RRHS Physics Page 9

Check Your LearningState the direction for each of the following vectors:a) 20 o S of Eb) 80 o W of Sc) 40 o W of Nd) 50 o N of We) Sf) 75 o E of NRRHS Physics Page 10

1.1.1 In Class or Homework Exercise1. Break the following vectors into components:a. 45 km in a direction 25 S of E;dysin 2545dy 19kmdxcos 2545dx 41kmb. 74 km, 35 E of Ndxsin 3574dx 42kmdycos3574dy 61km2. A bug crawled 34.0 cm E, then 48.5 cm S. What is the bug’s displacementfrom his starting point?In this question, you are given the components of the displacement vector.d34.0cmdxyd ?48.5cmRRHS Physics Page 11

dd (34.0) (48.5) 59.2cm2 248.5tan34.0 55.0The displacement isd 59.2 cm, 55.0 S of E3. A ship leaves its home port expecting to travel to a port 500. km due south.Before it can move, a severe storm comes up and blows the ship 100. km dueeast. How far is the ship from its destination? In what direction must the shiptravel to reach its destination?The ship must go 100. km west and 500. km south to reach its intendeddestination:dd (100.) (500.) 510. km2 2500.tan100. 78.7The ship must travel 510.km in a direction 78.7S of WRRHS Physics Page 12

1.1.2 Displacement Vectors and Vector AlgebraSince position and displacement vectors can be easily visualized and directlyrepresented using vector diagrams, it is these types of vectors that will be used inthis section to introduce vector addition and subtraction; the information here can,however, be applied to any kind of vector.Since vectors are not single numbers, our usual laws of algebra cannot be applied tothem; in other words, we cannot simply add the magnitude of two vectors together toobtain a total magnitude. Vector algebra requires the use of a vector diagram.Vector AdditionWhat does it mean to add two vectors? Consider displacement vectors a and b 3which represent successive displacements of a person walking.a bThe addition of these two displacements (the total displacement) should tell us wherethe person is at the end of his journey relative to where he started. We will use c torepresent the total displacement:c a bTo help visualize this, we must draw a vector addition diagram. Since the personcompleted displacement a and then completed displacement b , we draw the secondvector where the first one ends.3 To simplify notation, a and b are being used here instead of the standard drepresent displacement.with subscripts toRRHS Physics Page 13

aFigure 4: a bVectors can always be moved around (by sliding them) when drawing a vectordiagram, as long as their direction remains the same.The total displacement c (from where he started to where he ended) can then bedrawn.cbaFigure 5: a b cThis diagram represents the vector equation a b c . When we add two scalarstogether, the answer that we get is referred to as a sum. When we add two vectorstogether we get what is referred to as a resultant vector. So when we say thata b c , c is the resultant vector that goes from where we started to where weended.It is important to note that the vector equation a b c tells us how to draw thevector diagram shown in Figure 5, but that is all it should be used for. Numberscannot be substituted into a vector equation to be solved algebraically.When drawing a vector addition diagram, it is important to remember two things:1. The vectors being added must be drawn head to tail.2. The resultant vector goes from the tail of the first vector to the head of thesecond vector (from where you started to where you ended).RRHS Physics Page 14

Check Your LearningGiven the following vectors, draw a vector diagram to represent each of the followingvector equations (where x is an unknown vector in each case):abca. a b xb. b c xc. a b c 0d. a x be. x c axbcbcbaxabxxcaaRRHS Physics Page 15

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It is possible to obtain the magnitude and direction of the resultant from the diagramsthat were completed in the Check Your Learning by drawing scale diagrams with aruler and protractor; however, we will be using an algebraic approach to solve thistype of problem by looking at the components of the vectors. If we redraw Figure 4and add the components (as dotted lines) of each vector, we getabFigure 6: Vector AdditionRemember that the vector a actually represents the components ( a , a ) ; the othervector b represents the components ( b , b ) . If we add these two vectors, we arexactually adding their components. So a b will give ( a b , a b ) , and the diagramcan be redrawn to look like this:yx x y yxybaFigure 7: Vector AdditionThe only difference between these two diagrams is that the component vectors havebeen moved to show the x components together and the y components together –the original vectors a and b have not been changed. If we add the resultant vector cto the diagram we getacbFigure 8: Vector Addition with ResultantRRHS Physics Page 17

Notice that the vector c represents the sum of the components ( a b , a b ) . Wex x y ynow have one large right angle, so we can again use the Pythagorean theorem andour trigonometric functions to find the magnitude and direction of c .cFigure 9: Resultant Vector with ComponentsSince we now have a single right angle triangle, we can use the Pythagoreantheoremc ( x) ( y)2 2to find the magnitude of c and the angle can be found usingytan xExampleA person walks 5.0 km east and then 8.0 km in a direction 75N of E. What is hisdisplacement?Solutiond5.0km E1d 8.0 km,75Nof Ed2t?We know that the total displacement is the sum of the individual displacements,dt d1 d2This is a vector equation which will be used to draw the vector diagram:RRHS Physics Page 18

d td 2d 2 yd 1d 2 xThe vector d1is already in one of our 4 component directions (east), so we do notneed to break this vector up into components. We do have to break d2intocomponents, since it has both north and east components:d2xcos758.0d2.1km2xd2ysin 758.0d7.7km2 yd td 2Since we know that the resultant vector has an x-component of 7.1 km and a y-component of 7.7 km, we can now find the magnitude and direction:d d d2 2t tx ty(7.1) (7.7)10.km2 2using Pythagorean theoremandRRHS Physics Page 19

dtan dtytx7.77.147The total displacement isd10. km,47N of EtRRHS Physics Page 20

Check Your LearningA person walks 4.0 km south and then 7.2 km in a direction 21 o W of N. What wasthe person’s displacement?d 4.0 km, S 4.0km1d 7.2 km,21W of Nd2t?d d dt1 2d td 2d 1Since d1is south, it does not have to be broken into components; we do, however,have to break d2into components so that we can calculate the components of dt:d td 2 yd 2d 1d 2 xd2xcos697.2d2.6km2xd2ysin 697.2d6.7km2 yRRHS Physics Page 21

d txd td tyThe components of the displacement are simply the sum of the individualcomponents (be careful to indicate the signs as indicated by the direction of thearrows in the diagram):d d dtx 2x 1x 2.6 02.6kmd d dty 2y 1y 6.7 ( 4.0) 2.7kmNotice that since d1is directed south, there is no x component.d txd td tyd ( d ) ( d)2 2t tx ty(2.6) (2.7) 3.7km2 2dtan dtxty2.62.744Notice that the signs of the components do not have to be used when calculating themagnitude or the angle. The final displacement isd 3.7 km,44W of NtRRHS Physics Page 22

Vector SubtractionJust like subtraction of two scalars is really the same as adding a negative scalar (5 3 is the same as 5 ( 3) ), the subtraction of two vectors a b is the same asa( b); but ( b)just means ( b, b ) ; in other words, we are just changing thexydirection of the vector b and instead of adding the components of the two vectorswe subtract them. Using the same vectors as our previous example,a bFigure 10: Vectors a and ba b c is the same as a ( b) c so the two vectors that we must add are a and b :abFigure 11: Vectors a and bNotice that b is simply pointing in the direction opposite that of b . The vectordiagram for a ( b) c now looks likeacbFigure 12: a ( b) cRRHS Physics Page 23

The vector diagram can now be analyzed using components in the same way that itwas for the vector addition problems. The resultant vector c can also still berepresented in component forma xa yacbb yb xFigure 13: a ( b) c with componentswhere, in this case, x ax bxand y ay by.RRHS Physics Page 24

Check Your LearningGiven the following vectors, draw a vector diagram to represent each of the followingvector equations (where x is an unknown vector in each case):abca. a b xb. b c xc. a x bd. x c acxabbxbxaaxcRRHS Physics Page 25

1.1.2 In Class or Homework Exercise1. A delivery truck travels 18 blocks north, 16 blocks east, and 10 blocks south.What is its final displacement from the origin?The final displacement is the sum of all of the individual displacements:dd yd x 16blocksd x 18 ( 10)d y 8 blocksd d d2 2x y(16) (8)2 218blocksdtan dyx816 27d18blocks,27N of E2. A hiker leaves camp and, using a compass, walks 4 km E, 6 km S, 3 km E, 5km N, 10 km W, 8 km N, and 3 km S. At the end of three days, the hiker islost. Compute how far the hiker is from camp and which direction should betaken to get back to camp.This time we need to know the vector that goes from where the hiker ended towhere he started, so we are actually looking for the negative of thedisplacement vector.RRHS Physics Page 26

d 4 3 ( 10)d x3km ( 6) 5 8 ( 3)d y 4kmSo d 3kmand d 4kmxydd d d2 2t tx ty3 4 5km2 24tan3 50The hiker must go 5km in a direction 50 S of E to get back to camp.3. A car is driven 30.0 km west and then 80.0 km southwest. What is thedisplacement of the car from the point of origin (magnitude and direction)?The term southwest means that the direction is halfway between south andwest, so the angle is 45.d d dt1 2RRHS Physics Page 27

d 2 xd 1d 2 yd 2d tyd td txd2 x dcos(80.0)cos 45 56.6kmd2 y dsin(80.0)sin 45 56.6kmdty d2 y56.6kmd d dtx1 2x 30.0 ( 56.6)86.6kmd d d2 2t tx ty(86.6) (56.6)103km2 2dtan dtxty86.656.6 56.8d103 km, 56.8 W of S4. An explorer walks 22.0 km north, and then walks in a direction 60.0 south ofeast for 47.0 km.a. What distance has he travelled?d d dt1 222.0 47.0 69.0kmb. What is his displacement from the origin?d d dt1 2RRHS Physics Page 28

d 2 xd 2 yd tyd td txdcos2x d2 (47.0)(cos60.0 ) 23.5kmdsin2y d2 (47.0)(sin 60.0 ) 40.7kmd d dty1 2y 22.0 ( 40.7)18.7kmdtx d2x 23.5kmd d d2 2t tx ty(23.5) (18.7) 30.0km2 2dtan dtxty23.518.751.5d 30.0 km,51.5 E of Stc. What displacement vector must he follow to return to his originallocation?To get back to his original location, he must follow the negative of hisdisplacement vector:RRHS Physics Page 29

d tyd td txThe magnitude will still be the same; the angle will be90.0 51.5 38.5The vector that he must follow isd 30.0 km,38.5 N of W5. A man walks 3.0 km north, 4.5 km in a direction 40 . east of north, and 6.0km in a direction 60 . south of east. What is his displacement vector?d d d dt1 2 3d 2 xd 3xd 2 yd 2d 3d 3 yd 1d td tyd txd dsin 40. 2x2(4.5)sin 40. 2.9kmd dcos 40. 2y2(4.5)cos 40. 3.4kmdcos60.3x d3 (6.0)cos60. 3.0kmd dsin 60. 3y3(6.0)sin 60. 5.2kmRRHS Physics Page 30

d d dtx 2x 3x2.9 3.0 5.9kmd d d dty 1 2 y 3y 3.0 3.4 ( 5.2)1.2kmd d d2 2t tx ty(5.9) (1.2) 6.0km2 2dtan dtytx1.25.9 11d 6.0 km,11 N of Et6. After the end of a long day of travelling, Slimy the Slug is 255 cm east of hishome. If he started out the day by travelling 90.0 cm in a direction 25 o east ofnorth in the morning, how far did he travel in the afternoon (and in whatdirection) to get to his final location?dt12255cm Ed 90.0 cm,25 E of Nd ?d d dt1 2d d ( d)2 t1As can be seen, this problem can be solved as a vector subtraction problem:d td 2 yd 2d 1d 1yd 2 xd 1xRRHS Physics Page 31

d dsin 251x1(90.0)sin 25 38cmd dcos 251y1(90.0)cos 25 82cmd d ( d)2x t 1x 255 ( 38) 217cmd d2y1y82cmd d d2 22 2x2 y(217) (82) 232cm2 2dtan d2x2 y21782 69d2 232 cm,69 E of SRRHS Physics Page 32

1.1.3 Velocity VectorsYou saw in the previous section that an object's position in two dimensions is givenby two coordinates ( d , d ) . Remember from Unit 2 that velocity is the change inposition, or displacement, over time:dv txySince velocity is calculated using displacement and has the same direction as thedisplacement, velocity is also a vector which has two components ( v , v ) .xyvxdtxvydtyAs was discussed in Unit 2, there is no such thing as absolute velocity; the velocity ofan object is always relative to some frame of reference. Consider the example of adog on a boat. The boat is moving north at 7 m/s relative to the shore. Now supposethat the dog is moving north at 2 m/s relative to the boat. In other words, the dog ismoving 2 m/s faster than the boat. How fast is the dog actually moving? It dependson your point of view. To someone on the boat, the dog is moving at 2 m/s; however,to somebody on the shore, the dog is moving its 2 m/s plus the boat's 7 m/s (sincethey are moving in the same direction), which is 9 m/s.The situation is similar in two dimensions. Suppose that a boat is crossing a body ofwater at 5.0 m/s relative to the water (we will use the symbol vbwto represent thevelocity of the boat with respect to the water). 4 If the water is not moving, a personon the shore sees the boat moving at 5.0 m/s relative to the shore as well. Nowsuppose that the body of water is a river flowing perpendicular to the boat at 3.5 m/sas measured by someone on the shore ( v ws).The person on the shore now sees the river carrying the boat downstream at 3.5 m/s,but also sees the boat moving across the river at 5.0 m/s. Just like the dog on theboat, the person on the shore sees the addition of the two velocities, so the velocityof the boat with respect to the shore is given by4 Using this notation, the first subscript identifies the object that is moving, the second subscriptidentifies the frame of reference with respect to which it is movingRRHS Physics Page 33

vbs vbw vwsRemember, however, that these quantities are vectors and must therefore be addedas vectors! (as was described in the last section – in other words, we must draw avector diagram).By using subscripts according to the convention described above we see that theinner subscripts on the right-hand side of the equation are the same as each other(in this case, w) and the outer subscripts on the right-hand side of the equation arethe same as the subscripts for the resultant vector on the left vbs. This can be usedas a check if you are not sure if you are adding the proper vectors.Since they are vectors, remember, these velocities must be added as vectors – theymust be drawn head to tail:v wsv bwv bsFigure 14: vbs vbw vwsThe resultant vector (the velocity actually observed by someone on the shore) is thevector vbs. This resultant velocity has two components (one across the river and onedown the river). Note that the component across the river is the same as the originalvelocity of the boat that was directed across the river; therefore, the boat will crossthe river in the same amount of time with the river flowing as without!ExampleA boat that has a speed of 5.0 m/s in still water heads north directly across a riverthat is 250 m wide. The velocity of the river is 3.5 m/s east.a. What is the velocity of the boat with respect to the shore?b. How long does it take the boat to cross the river?c. How far downstream does the boat land?d. What heading (direction) would the boat need in order to land directly acrossfrom its starting point?Solutiona.RRHS Physics Page 34

vvvbwwsbs 5.0 m / s N 3.5 m / s E ?To find the velocity of the boat with respect to the shore, we must add thevelocity of the boat with respect to the water and the velocity of the water withrespect to the shore:vbs vbw vwsv ws3.5 m/sv bwv bs5.0 m/sv bsUsing the Pythagorean Theorem,orv v v2 2bs bw ws(5.0) (3.5)2 2 6.1 m/sSince we are finding velocity, we must also find the direction3.5tan 5.035vbs6.1 m / s,35 E of Nb. Since the boat is going at a constant velocity, we can usedv tNotice that both the displacement and velocity are vectors, and must have thesame direction; therefore, if we use the displacement of 250 m (which is north)than we must also use the component of the velocity that is north, which is theboat’s velocity with respect to the water.RRHS Physics Page 35

d250m Nvbw 5.0 m / s Nt ?dv t2505.0 tt 50. sc. Similarly, if we want to know how far downstream (east) the boat travels wemust use the component of the velocity in this direction. The boat will betravelling for the same amount of time downriver that it took to cross the river.vws 3.5 m / s Et 50. sd?dv td3.5 50.d 180md. If the boat is to land directly across from its starting point, than the velocity ofthe boat with respect to the shore must be north.v v vbs bw wsvUsing this equation, and knowing that the resultant vector bs must be north,our vector diagram will look like thisv wsv bwv bsRRHS Physics Page 36

vsinvwsbw3.55.0 44The boat must head 44 W of N. Notice that this is not the same as the anglethat the boat would travel downstream if no corrective action is taken.RRHS Physics Page 37

Check Your LearningIn the example just completed, how long will it take the boat to cross the river in part(d)? In other words, how long will it take to cross the river when corrective action istaken so that the boat lands directly across from its starting point? How does thisanswer compare with the time obtained in part (a) of the example?Since we know that it is 250 m across the river (north), we must find the boat’s newvelocity in this direction ( v bs).v wsv bwv bsvtan vwsbs3.5tan 44vvbsbs 3.6 m / s Ndv t2503.6 tt 69sIt takes more time to cross the river when correcting for the flow of the river than wascalculated in part (a) since some of the boat’s velocity is being used to compensatefor the river and stop the boat from moving downstream.RRHS Physics Page 38

1.1.3 In Class or Homework Exercise1. An airplane is travelling 1000. km/h in a direction 37 o east of north.a. Find the components of the velocity vector.vvxvsin 37(1000.)sin 37 600 km / hvyvcos37(1000.)cos37 800 km / hb. How far north and how far east has the plane travelled after 2.0 hours?vxxdx600 2.0dxdt 1200 km eastvyydy800 2.0dydt 1600 km north2. A motorboat whose speed in still water is 8.25 m/s must aim upstream at anangle of 25.5 o (with respect to a line perpendicular to the shore) in order totravel directly across the stream.a. What is the speed of the current?v 8.25 m / s,25.5upstream from the perpendicularbwv ?wsvbs vbw vwsv wsv bwv bsRRHS Physics Page 39

vsin vwsbwvwssin 25.58.25vws 3.55 m / sb. What is the resultant speed of the boat with respect to the shore?v 8.25 m / s,25.5upstream from the perpendicularbwv ?bsUsing the above diagram,vbscos vbwvbscos 25.5 8.25vbs 7.45 m / s3. Kyle wishes to fly to a point 450 km due south in 3.0 h. A wind is blowing fromthe west at 50.0 km/h. Compute the proper heading and airspeed that Kylemust choose in order to reach his destination on time.Let vag(the velocity of the air with respect to the ground) be the velocity of thewind; vpawill be the velocity of the plane with respect to the air (which will givethe heading and required air speed).dyt 3.0hvvagpa450km S 50.0 km / h E ?Since we know where he wants to be in 3.00 h, we can compute the requiredgroundspeed (speed of the plane with respect to the ground)vpgdyt450km3.0150km / h SRRHS Physics Page 40

vpg vpa vag(In other words, the plane’s air velocity added to the wind willgive us the ground velocity)v pav pgv agv v v2 2pa pg ag150 502 2160 km / hvtanvagpg50.0150 18vpa160 km / h,18 W of S4. Diane aims a boat that has a speed of 8.0 m/s in still water directly across ariver that flows at 6.0 m/s.a. What is the resultant velocity of the boat with respect to the shore?vvvbwwsbs8.0 m / s 6.0 m / s ?vbs vbw vwsv wsv bwv bsRRHS Physics Page 41

v v v2 2bs bw ws(8.0) (6.0)10. m/s2 26.0tan 8.037vbw10. m / s,37 downstream from the perpendicularb. If the stream is 240 m wide, how long will it take Diane to row across?d240mvbw 8.0 m / st ?dv t2408.0 tt 30. sc. How far downstream will Diane be?v 6.0 m / swst 30. sd?dv td6.0 30.d 180m5. An airplane whose airspeed is 200. km/h heads due north. But a 100. km/hwind from the northeast suddenly begins to blow. What is the resulting velocityof the plane with respect to the ground?v 200. km / h Nvvpaagpg100. km / h SW ?v v vpg pa agRRHS Physics Page 42

v agxv agyv agv pav pgv pgyv pgxvagyvagsin 45(100.)sin 45 70.7 km / hvagxvagcos 45(100.)cos 45 70.7 km / hv v vpgy pa agy 200. ( 70.7)129 km / hvpgx vagx 70.7 km / hv v v2 2pg pgx pgy(70.7) (129)147 km / h2 2vtan vpgypgx12970.761.3vpg147 km / h,61.3 N of W6. An airplane is heading due north at with an airspeed of 300. km/h. If a windbegins blowing from the southwest at a speed of 50.0 km/h, calculatea. the velocity of the plane with respect to the ground, andv 300. km / h Nvvpaagpg 50.0 km / h NE ?vpg vpa vagRRHS Physics Page 43

v agxv agyv agv pgv pgyv pav pgxvvagxpgxvagsin 45(50.0)sin 45(50.0)cos 45 35.4 km / h vagxvagyvagcos 45 35.4 km / hv v vpgy pa agy 35.4 km / h300. 35.4 335 km / hv v v2 2pg pgx pgy(35.4) (335) 337 km / h2 2vtan vpgypgx33535.484.0vpg337 km / h,84.0 N of Eb. how far off course it will be after 30.0 min if the pilot takes no correctiveaction.It is only the east component of the wind that is blowing the plane offcourse, since the plane set out going north; the north component isactually helping the plane go north.RRHS Physics Page 44

dv td35.4 0.500d 17.7kmc. Assuming that the pilot has the same airspeed of 300. km/h, whatheading should he use to maintain a course due north?We still have the same vector equation as before, but now the velocityof the plane with respect to the ground must be pointed north.v v vpg pa agv agxv agyv agv pgv payv pav paxvpaxvagx35.4 km / hvcos vpaxpa35.4300.83.283.2 N of Wd. What is his new speed with respect to the ground?RRHS Physics Page 45

vsin vpaypavpaysin 83.2300.v 298 km / hpayv v vpg pay agy298 35.4 333 km / h7. A swimmer is capable of swimming 1.80 m/s in still water.a. If she aims her body directly across a 200.0 m wide river whose currentis 0.80 m/s, how far downstream (from a point opposite her startingpoint) will she land?v v 0.80 m / sxwsv v 1.80 m / syddyxpw200.0m?First, find out how long it takes to cross the river:dyvyt200.01.80 tt 111sThis time can now be used to calculate how far downstream she iscarried by the water:vxxdx0.80 111dxdt 89mb. What is her velocity with respect to the shore?v v vps pw wsv wsv pwv psRRHS Physics Page 46

v v v2 2ps pw ps(1.80) (0.80) 2.0 m/s2 20.80tan 1.8024vbw2.0 m / s,24 downstream from the perpendicularc. At what upstream angle must the swimmer aim if she is to arrive at apoint directly across the stream?The velocity of the boat with respect to the shore must now be in the y-direction:v wsv pwv psvsinvwspw0.801.80 26The swimmer must aim 26 upstream from the perpendicular8. A pilot wishes to make a flight of 300. km northeast in 45 minutes. If there is tobe an 80.0 km/h wind from the north for the entire trip, what heading andairspeed must she use for the flight?dpg300.km NEt 45min 0.75hvvvagpapg 80.0 km / h S ?dpgt300.0.75 400 km / hv v vpg pa agv v ( v)pa pg agRRHS Physics Page 47

v paxv agv payv pav pgv pgyv pgxvpgyvpgcos 45vpgxvpgsin 45(400)cos 45 280 km / h(400)sin 45 280 km / hvpax vpgx 280 km / hv v ( v)pay pgy ag280 80.0 360 km / hv v v2 2pa pax pay(280) (360) 460 km / h2 2460 km / h,38 E of Nvtan vpaxpay280360389. A car travelling at 15 m/s N executes a gradual turn, so that it then moves at18 m/s E. What is the car's change in velocity?vi15 m / s Nv vf vivf18 m / s E vf ( vi)v?RRHS Physics Page 48

v fvv iv v v2 2f i(18) (15) 23 m/s2 2vitan vf151840.v 23 m / s,40. S of E10. A plane's velocity changes from 200. km/h N to 300. km/h 30.0 W of N. Findthe change in velocity.vi 200. km / h Nv vf vivf300. km / h,30.0 W of N vf ( vi)v?v fxv iv fyv fv yvvfxvfv xsin 30.0(300.)sin 30.0150. km / hvfyvfcos30.0(300.)cos30.0 260. km / hRRHS Physics Page 49

vxvfx150. km / hv v ( v)y fy i 260. ( 200.) 60. km / hv v v2 2x y(150) (60.)160. km / h2 2vtan vyx60.150.22v160 km / h,22 N of W11. A plane is flying at 100. m/s E. The pilot changes its velocity by 30.0 m/s in adirection 30.0 N of E. What is the plane's final velocity?vi100. m / s Ev vf viv 30.0 m / s,30.0 N of Evf vi vv ?fv fxv fyv fvv yv iv xv vcos30.0x(30.0)cos30.0 26.0 m/sv vsin 30.0y(30.0)sin 30.015.0 m/svfyvy15.0 m/sv v vfx i x100. 26.0126 m/sv v v2 2f fx fy(126) (15.0)127 m/s2 2vtanvfxfy12615.0 83.2vf127 m / s, 83.2 E of NRRHS Physics Page 50

12. A ferryboat, whose speed in still water is 2.85 m/s, must cross a 260 m wideriver and arrive at a point 110 m upstream from where it starts. To do so, thepilot must head the boat at a 45 o upstream angle. What is the speed of theriver's current?v 2.85 m / sbwddvwsbsybsx ?260m110md bsxd bsd bsydtan dbsxbsy110260 23v wsv bsxvbwyv bwv bsvbsyv bwxvbwxvbwcos 45(2.85)cos 45 2.0 m/svbwyvbwsin 45(2.85)sin 45 2.0 m/svbsy vbwy 2.0 m/svbsx vbsytan(2.0) tan 23 0.85 m/sRRHS Physics Page 51

v v vbsx bwx ws0.85 2.0 vvws 1.2 m / swsRRHS Physics Page 52

1.1.4 Module SummaryIn this module you have learned:o To represent two dimensional vectors using two methods:• components• magnitude and directiono How to add and subtract displacement vectors using vector diagrams andcomponents:• When adding vectors, they must be drawn head to tail. The resultantvector goes from the tail of the first vector to the head of the last vector.• Subtracting vectors is the same operation as adding a negative vector,where a negative vector points in the direction opposite the direction ofthe original vector.o How to use vector diagrams and components to calculate relative velocities.RRHS Physics Page 53

Module 1.2 Force Vectors1.2.1 Dynamics Problems in 2DIn Unit 3, you solved many problems involving horizontal and vertical forces byapplying Newton's 2 nd Law to different situations using free body diagrams. This willnow be extended to situations where the forces are no longer solely in the horizontalor vertical directions. Remember that Newton's 2 nd Law ( Fnet ma ) is a vectorequation, since it states the relationship between acceleration and net force, both ofwhich are vectors. This means that the acceleration and the net force will be in thesame direction; therefore, if we want to use scalar algebra to solve a problem, wemust use this equation in only one dimension at a time ( x or y ).In the diagram below, a man is pulling a box with a rope that makes an angle withthe ground.Note that the expected acceleration (horizontal) for this box and the applied force areneither parallel nor perpendicular, so Newton's 2 nd Law cannot be applied yet. A freebody diagram for this box would look like this:Notice that although the normal, friction, and gravity forces are all solely in the x(horizontal) or y (vertical) directions, the force of the man pulling is not. This can befixed if we break this force up into its x and y components.RRHS Physics Page 54

In other words, pulling with a forceseparate forces ofFxto the right andFpat an angle of is the same as applying twoFyupward. As can be seen in the diagramabove, all of the forces are now either in the x or y direction if we replace Fpwithits components Fpxand Fpy. We can now analyze the forces in each dimensionusing Newton's 2 ndLaw.Vertical: We will take up as the positive direction; therefore, FNandpositive andFpywill both beFgwill be negative. Remember, only vertical forces can be used tocalculate a vertical acceleration.andmayyFma F F FNypymay FN Fpy FgF0 FN Fpy Fg FFN g pygsince the vertical acceleration is zero (the box is not accelerating up or down). Noticethat FN Fg. Because we often know Fgand Fpy, we can solve for FNand use it inour calculation ofFf(remember that Ff FN, where is the coefficient of friction).Horizontal: We will take the right to be positive; therefore,Fpxwill be positive andFfwill be negative. Remember, only horizontal forces can be used to calculate ahorizontal acceleration.mamaxxFx F Fpxma F Fx px ffRRHS Physics Page 55

This can then be used with the horizontal acceleration. These are not equations tobe memorized and applied to all problems!!! This is a sample analysis of a typicalfree body diagram involving forces at an angle. As always when dealing with forceproblems, analysis should always start with a free body diagram.ExampleA 52.0 kg sled is being pulled along a frictionless horizontal ice surface by a personpulling a rope with a force of 235 N. The rope makes an angle of 35.0 o with thehorizontal. What is the acceleration of the sled?SolutionFirst we need a free body diagram:m 52.0kgFp 235N 35.0ax ?We must first calculate the components for the applied force:Fpx F cospFpy F sinp(235)cos35.0(235)sin 35.0193N135NHorizontal:RRHS Physics Page 56

mama(52.0) a 193axxxx FpxFx 3.71 m / s2Since there was no friction involved in this problem, we did not need a calculation ofthe normal force; since there was no vertical acceleration and the normal force wasnot needed, we were not required to do an analysis of the vertical forces.RRHS Physics Page 57

Check Your LearningA 52.0 kg sled is being pulled along a horizontal surface by a person pulling a ropewith a force of 235 N. The rope makes an angle of 35.0 o with the horizontal. If thecoefficient of friction between the sled and the surface is 0.25, what is theacceleration of the sled?Notice that this problem is the same as the previous example except for the fact thatwe now have friction involved.First we need a free body diagram:m 52.0kgFp 235N 35.0 0.25a ?xWe must first calculate the components for the applied force:Fpx F cospFpy F sinp(235)cos35.0(235)sin 35.0193N135NSince we need the force of friction, we will need the normal force; therefore, it isnecessary to look at the vertical forces:RRHS Physics Page 58

Vertical:Using up as positive,mayFma F F F0 F F Fyy N g pyF F FN g pyN g py(52.0)(9.80) 135 375NNotice that the normal force is not equal in magnitude to the force of gravity!Horizontal:Ff FN (0.25)(375) 94Nmax(52.0) a 193 94xx px fx px fxxFma F Fma F Fa 1.9 m / s2RRHS Physics Page 59

1.2.1 In Class or Homework Exercise1. A 25.0 kg sled is accelerating at 2.3 m/s 2 . A force of 300.0 N is pulling the sledalong a rope that is being held at an angle of 35 o with the horizontal. What isthe coefficient of friction?m 25.0kgFp 300.0N 35 ?a 2.3 m / sx2Fg mg (25.0)(9.80) 245NWe must first calculate the components for the applied force:Fpx F cosp(300.0)cos35 246NFpy F sinp(300.0)sin 35172NVertical:Using up as positive,mayFma F F F0 F F Fyy N g pyF F FN g pyN g py245 172 73NHorizontal:RRHS Physics Page 60

max(25.0)(2.3) 246Fxx px fx px ffFma F Fma F FF189NfFf FN189 (73) 2.62. A 15.0 kg sled is being pulled along a horizontal surface by a rope that is heldat a 20.0 o angle with the horizontal. The tension in the rope is 110.0 N. If thecoefficient of friction is 0.30, what is the acceleration of the sled?m 15.0kgFp 110.0N 20.0 0.30a ?xFg mg (15.0)(9.80)147NWe must first calculate the components for the applied force:Fpx F cosp(110.0)cos 20.0103NFpy F sinp(110.0)sin 20.0 37.6NRRHS Physics Page 61

Vertical:Using up as positive,mayFma F F F0 F F F147 37.6109NHorizontal:F Fyy N g pyF F FfN g pyN g pyN (0.30)(109) 33Nmax(15.0) a 103 33xx px fx px fxxFma F Fma F Fa 4.7 m / s23. A man pushes a 15 kg lawnmower at constant speed with a force of 90.0 Ndirected downward along the handle, which is at an angle of 30.0 o to thehorizontal. What is the coefficient of friction?m 15kgFp 90.0N 30.0Fg mg (15)(9.80)147N ?RRHS Physics Page 62a 0xa 0y

We must first calculate the components for the applied force:Fpx F cospFpy F sinp(90.0)cos30.0(90.0)sin 30.0 77.9N 45.0NVertical:Using up as positive,mayFma F F F0 F F F147 45.0192Nyy N g pyF F FN g pyN g pyHorizontal:maxxx px fx px f0 77.9FfFma F Fma F FF 77.9NfFf FN77.9 (192)0.4064. A sled is being pulled by a rope that makes an angle of 27 o with thehorizontal. The coefficient of friction between the sled and the ground is 0.30.If the tension in the rope is 245 N and the acceleration of the sled is 1.2 m/s 2 ,what is the mass of the sled?RRHS Physics Page 63

m ?F 245N 27 0.30apx 1.2 m / sa 0y2We must first calculate the components for the applied force:Fpx F cosp(245)cos 27 220NFpy F sinp(245)sin 27110NVertical:Using up as positive,mayFma F F F0 F F F9.80m110Horizontal:F Fyy N g pyF F FfN g pyN g pyN(0.30)(9.80m110)2.9m33RRHS Physics Page 64

maxFma F Fm(1.2) 220 (2.9m33)1.2m 220 2.9m334.1m 253m 62kgxx px fma F Fx px f5. A 55.0 kg rock is being pulled along a horizontal surface at a constant speed.The coefficient of friction between the rock and the surface is 0.76. If the ropepulling the rock is at a 40.0 o angle with the horizontal, with what force is therock being pulled?m 55.0kgF ? 40.0 0.76apx 0a 0yFg mg (55.0)(9.80) 539NWe must first calculate the components for the applied force:Fpx F cosp F cos 40.0p 0.766FpFpy F sinp F sin 40.0p 0.643FpVertical:Using up as positive,RRHS Physics Page 65

mayFma F F F0 F F Fyy N g pyF F FN g pyN g py539 0.643FpHorizontal:Ff FN0.76(539 0.643 F )410 0.49Fppmaxxx px fx px f0 0.766 F (410. 0.49 F )0 0.766F 410. 0.49F410 1.26FpFma F Fma F FFp 330Npppp6. A 40.0 kg iceboat is gliding across a frozen lake with a constant velocity of 14m/s E, when a gust of wind from the southwest exerts a constant force of 100.N on its sails for 3.0 s. With what velocity will the boat be moving after thewind has subsided? Ignore any frictional forces.m 40.0kgv 14 m / s Eit 3.0sF 100.N NEvf ?RRHS Physics Page 66

Fnet ma100. (40.0) a2a 2.50 m / s NEva tv2.50 3.0v7.5 m / s NEv v vv v vfifiv fxvfyv fvvyv iv xv v 7.5cos45 5.3 m / sxyv v vfx i x14 5.319 m/svfyvy 5.3 m/sv v v2 2f fx fy(19) (5.3) 20. m/s2 2vtanvfxfy195.3 74vf20. m / s, 74 E of N7. Two tow trucks attach ropes to a stranded vehicle. The first tow truck pullswith a force of 25000 N, while the second truck pulls with a force of 15000 N.The two ropes make an angle of 15.5 o with each other. Find the resultantforce on the vehicle.This problem can be simplified by aligning one of the forces along the x-axisRRHS Physics Page 67

F 25000N1FF2net15000N ?F F Fnet1 2F cos15.52x F2 (15000)cos15.514000NF F Fnetx1 2x25000 14000 39000NFnetyF sin15.52y F2 (15000)sin15.5 4000N F2 y 4000NF ( F ) ( F )2 2net netx nety(39000) (4000) 39000N2 2FtanFnetynetx400039000 5.9Fnet39000 N,5.9from the larger forceRRHS Physics Page 68

1.2.2 Inclined PlanesUp until now, we have dealt only with dynamics situations where the motion hasbeen either horizontal or vertical, and where we have calculated all of thecomponents of our vectors in horizontal and vertical planes. We are now going toapply force vectors and Newton's second law to an inclined plane (a ramp). On aninclined plane, the acceleration is neither horizontal nor vertical. It is parallel to theplane.If we place a box on a ramp (ignoring friction for now), as in Figure 15,Figure 15: Rampit can be observed that there are only two forces acting on the box - the normal forceFN(which is perpendicular to the surface of the plane) and the force of gravityDrawing a free body diagram, we getFg.Figure 16: Ramp Free Body DiagramNotice that these vectors exist in two dimensions and are not in component form(they are not either parallel or perpendicular to one another). In order to applyNewton's second law, we want to analyze the forces one dimension at a time.Instead of using our usual coordinate system containing horizontal and vertical axes,RRHS Physics Page 69

it makes more sense in this situation to rotate our axes so that they areperpendicular and parallel to the surface of the inclined plane (the same direction asthe acceleration). In other words, our x direction will be parallel to the plane and they direction will by perpendicular to the plane.Since the normal force is already perpendicular to the plane (and any applied forcealong with the force of friction will usually be parallel to the plane), only the force ofgravity must be broken up into components.Figure 17: Ramp Free Body Diagram with ComponentsThis can be done as shown in Figure 18 (where thediagram has been enlarged).Fgvector from the previousFigure 18: Gravity ComponentsThe angle in the top of the triangle is the same angle as the slope of the inclinedplane (try showing this using geometry). Using trigonometry, it can be found that thetwo components areandF mg sin(1.1)gxF mg cos(1.2)gyRRHS Physics Page 70

The two components of gravity can be thought of as having different roles in thissituation: Fgxis down (and parallel) to the ramp and is trying to accelerate the boxdown the ramp; Fgyis pushing the box onto the ramp and is responsible for keeping itin contact with the ramp.If we apply Newton’s Second Law to the perpendicular forces we getmamayyFNy F Fgymay FN FgySince there is no acceleration perpendicular to the plane we now havem F F(0)N gyFN Fwhere Fgycan be found using equation (1.2). If friction is present, the normal forcecan then be used in this calculation. Again notice that FN Fg.gySimilarly, the parallel forces can be used to obtain an expression for the parallelacceleration on the inclined planewheremamaxx F FgxF can be found using equation (1.1). Notice that this is just a simplegxanalysis where friction and other external forces have not been included; if present,these would have to be considered in the force analysis. Again, it is extremelyimportant to draw a free body diagram at the start of the problem!xExampleA 1200 kg car is on an icy (frictionless) hill that is inclined at an angle of 12 o with thehorizontal. What is the acceleration of the car down the hill?SolutionSince there is no friction, the only forces involved are gravity and the normal force:RRHS Physics Page 71

Since the acceleration that we are looking for is parallel to the inclined plane, we willrotate our axes so that the x-axis is parallel to the inclined plane.m 1200kg 12aayx 0 ?Parallel ForcesIf we use down the ramp as positive,mamamaaxxxx FgxFx mg sin(9.80)(sin12 ) 2.0 m/s2Although we were given the mass of the car, notice that it cancelled out in thisproblem and was not needed.RRHS Physics Page 72

Check Your LearningA 1200 kg car is on an icy hill that is inclined at an angle of 12 o with the horizontal.As the car starts sliding, the driver locks the wheels. The coefficient of frictionbetween the locked wheels and the icy surface is 0.14. What is the acceleration ofthe car down the hill?Adding friction to our free body diagram we getSince the acceleration that we are looking for is parallel to the inclined plane, we willagain rotate our axes so that the x-axis is parallel to the inclined plane.m 1200kg 12 0.14aayx 0 ?First, we will find the components of the force of gravity:Fgx mg sin(1200)(9.80)sin12Fgy mg cos(1200)(9.80)cos12 2450N11500NSince there is friction this time, we will need the normal force. We must thereforelook at the perpendicular forces:Perpendicular ForcesRRHS Physics Page 73

may0 F 11500yy N gyy N gyNNFma F Fma F FF 11500Parallel Forces (using down the ramp as positive)Ff FN (0.14)(11500)1610Nmax(1200) a 2450 1610xx gx fx gx fxxFma F Fma F Fa 0.70 m / s2It could also be shown that the mass will again cancel out in this problem and wasnot needed.RRHS Physics Page 74

1.2.2 In Class or Homework Exercise1. An 18.0 kg box is released on a 33.0 o incline and accelerates at 0.300 m/s 2 .What is the coefficient of friction?m 18.0kg 33.0 ?aayx 0 0.300 m / s2First, we will find the components of the force of gravity:Fgx mg sin(18.0)(9.80)sin 33.0 96.1NFgy mg cos(18.0)(9.80)cos33.0148NPerpendicular Forces (using away from the ramp as positive)ma Fyma F F0 F 148yy N gyma F FFy N gyNN 148NParallel Forces (using down the ramp as positive)RRHS Physics Page 75

ma(18.0)(0.300) 96.1FFf Fxxx gx fx gx ffFma F Fma F FNF90.7 (148) 90.7Nf0.6132. A box (mass is 455 g) is lying on a hill which has an inclination of 15.0 o withthe horizontal.a. Ignoring friction, what is the acceleration of the box down the hill?m 455g 0.455kg 15.0aayx 0 ?First, we will find the components of the force of gravity:Fgx mg sinFgy mg cos(0.455)(9.80)sin15.0(0.455)(9.80)cos15.01.15N 4.31NSince there is no friction this time, we do not need to look at theperpendicular forces:Parallel Forces (using down the ramp as positive)RRHS Physics Page 76

mamama(0.455) a 1.15axxxxx F FgxgxFx 2.53 m / s2b. If there is a coefficient of friction of 0.20, will the box slide down the hill?If so, at what acceleration?m 455g 0.455kgFFgxgy 1.15N 0.86N 0.20aayx 0 ?Since there is friction this time, we will need the normal force. We musttherefore look at the perpendicular forces:Perpendicular Forcesma Fyma F F0 F 4.31yy N gyma F FFy N gyNN 4.31NParallel Forces (using down the ramp as positive)RRHS Physics Page 77

Ff FN (0.20)(4.31) 0.86NSince Fgx Ff, the box will accelerate down the hill.max(0.455) a 1.15 0.86xx gx fx gx fxxFma F Fma F Fa 0.64 m / s2c. How much force is required to push the box up the ramp at a constantspeed?Since the box is now going up the hill, friction must be down the hill:m 0.455kgFFaafgxxy 0.86N 1.15N 0 0Using up the ramp as positive and looking at the parallel forces,RRHS Physics Page 78

maxFma F F F0 F F F1.15 0.86xx gx f pma F F Fx gx f pp gx fF F Fp gx f2.01N3. A 165 kg piano is on a 25.0 ramp. The coefficient of friction is 0.30. Jack isresponsible for seeing that nobody is killed by a runaway piano.a. How much force (and in what direction) must Jack exert so that thepiano descends at a constant speed?Initially, we do not know what direction Jack has to apply his force sothat the piano descends at a constant speed. If we look at the FreeBody Diagram without Jack’s forcem 165kg 25.0 0.30aaFxyp 0 0 ?We see that we must compare FgxandFgx mg sin(165)(9.80)sin 25.0 683NFgyFf mg cos(165)(9.80)cos 25.01470NPerpendicular Forces:RRHS Physics Page 79

may0 F 1470yy N gyy N gyNNFma F Fma F FF 1470NFf FN (0.30)(1470) 440NSince Fgx Ff, the piano will accelerate down the ramp if Jack doesnothing; therefore Jack must push up the ramp:Using up the ramp as positive,Parallel Forces:maxFma F F F0 F F Fxx p f gxp f gxF F Fp gx f683 440 240NThe force must be directed up the ramp.b. How much force (and in what direction) must Jack exert so that thepiano ascends at a constant speed?RRHS Physics Page 80

FFaFgxfxp 683N 440N 0 ?Using up the ramp as positive,maxFma F F F0 F F Fxx p f gxp f gxF F Fp gx f683 440 1120NThe force must be directed up the ramp.4. A car can decelerate at -5.5 m/s 2 when coming to rest on a level road. Whatwould the deceleration be if the road inclines 15 o uphill?The coefficient of friction will be the same in each situation (same surfaces).We must find this from the level surface part of the problem.Level Roada 5.5 m / s ?2Using the forward direction as positive,RRHS Physics Page 81

ma Fma Ffma FNma mg5.5 (9.80) 0.56Incline 0.56 15aayx 0 ?Since there is no perpendicular acceleration, FN Fgy.Looking at the parallel forces and using up the ramp as positive,maxxx gx fx gx fma mg sin mg cosxxFma F Fma F Fa (9.80)sin15 (0.56)(9.80)cos157.8 m/s25. A sled is sliding down a hill and is 225 m from the bottom. It takes 13.5 s forthe sled to reach the bottom. If the sled’s speed at this location is 6.0 m/s andthe slope of the hill is 30.0 , what is the coefficient of friction between the hilland the sled?Using down the ramp as positive,RRHS Physics Page 82

d225mv 6.0 m / sit 13.5s 30.0 ?d v t a t1x i 2 x225 (6.0)(13.5) a (13.5)ax1.6 m / s2212x2maxma mg sin mg cos1.6 (9.80)sin 30.0 (9.80)cos30.00.39xx gx fx gx fx gx NxFma F Fma F Fma F F 6. A 5.0 kg mass is on a ramp that is inclined at 30 o with the horizontal. A ropeattached to the 5.0 kg block goes up the ramp and over a pulley, where it isattached to a 4.2 kg block that is hanging in mid air. The coefficient of frictionbetween the 5.0 kg block and the ramp is 0.10. What is the acceleration of thissystem?Figure 19: Diagram for Question 6RRHS Physics Page 83

m 5.0kgm12 4.2kg 0.10a ?We must first determine the direction of acceleration of the system bycomparing Fg 2and Fg1x.Fg1xm1gsin(5.0)(9.80)sin 30.0 24.5NFg m g2 2 (4.2)(9.80) 41.2NSince Fg2 Fg 1x, the system will accelerate clockwise and friction on the 5.0kg object will be down the ramp.If we treat the whole system as one object (as in Unit 3), our linearized freebody diagram will look like this:The tension in the rope can be ignored since we are treating the system asone big object.Ff FN Fgy mgcos(0.10)(5.0)(9.80)cos30.0 4.2NUsing clockwise as the positive direction (toward the 4.2 kg mass),RRHS Physics Page 84

ma Fm a F F Ft g 2 g1x fm a F F Ft g 2 g1x f(9.2) a 41.2 24.5 4.2a 1.4 m / s27. A force of 500.0 N applied to a rope held at 30.0 o above the surface of a rampis required to pull a box weighing 1000.0 N at a constant velocity up the plane.The ramp has a base of 14.0 m and a length of 15.0 m. What is the coefficientof friction?Figure 20: Diagram for Question 7Fp 500.0N 30.0Fg 1000.0Na 0axy 0 ?Fg mg1000.0 m(9.80)m 102kgRRHS Physics Page 85

We must break the applied force up into components that are parallel andperpendicular to the ramp.F F cosF F sinpxp(500.0)(cos30.0 )(500.0)(sin 30.0 ) 433N 250. NpypNext we must find the angle that the ramp makes with the horizontal so thatwe can find the components of the force of gravity:14cos15 21Fgx mg sin(102)(9.80)sin 21 360NFgy mg cos(102)(9.80)cos 21 930NPerpendicular Forcesmayy py N gy0 F F Fypy N gy0 250 F 930NFma F F FF 680NNParallel Forces (using up the ramp as positive)maxx px gx f0 F F F0 433 360 FfFma F F FF 73Nxpx gx ffFf FN73 (680) 0.118. If a bicyclist (75 kg) can coast down a 5.6 o hill at a steady speed of 7.0 km/h,how much force must be applied to climb the hill at the same speed?We can assume the same magnitude for the force of friction on the way upand the way down the hill:Down the HillRRHS Physics Page 86

m 75kg 5.6a 0v 7.0 km / hFf ?Parallel Forces (Using down the ramp as positive)max0 Fxx gx fgxFma F FFFf mg sin(75)(9.80)sin(5.6 ) 72NUp the Hillm 75kg 5.6a 0v 7.0 km / hFfp 72NF ?fRRHS Physics Page 87

Parallel Forces (Using down the ramp as positive)maxFma F F F0 F F Fxx gx f pgx f pF F Fp gx f72 72 144NRRHS Physics Page 88

1.2.3 Module SummaryIn this module you learned thato Force vectors can be broken into components so that dynamics situations canbe analyzed using free body diagrams and Newton’s Second Law.o Situations involving inclined planes (ramps) can be analyzed by rotating thecoordinate system so that the x-axis is parallel to the ramp and the y-axis isperpendicular to the ramp. The components for the force of gravity can thenbe given byF mg sinFgxgy mg cosRRHS Physics Page 89

Module 1.3 Equilibrium1.3.1 Translational EquilibriumYou saw in Unit 3 that if two equal but opposite forces are applied to an object, thenet force is zero and the object is said to be in equilibrium. A body in equilibrium atrest in a particular reference frame is said to be in static equilibrium; a body movinguniformly at constant velocity is in dynamic equilibrium. We will be dealing withmainly static equilibrium in this module, although the net force is zero in both cases.We will now extend our discussion of equilibrium to two dimensions.The type of equilibrium discussed in Unit 3 is known as translational equilibrium.Consider a mass being supported in midair by two ropes.Figure 21: Hanging MassThe mass is stationary; therefore, it is obviously not accelerating either sideways orup and down. The net force must therefore be zero and the object is said to be intranslational equilibrium.RRHS Physics Page 90

Figure 22: Hanging Mass Free Body DiagramAs can be seen by the free-body diagram, there are three forces acting on the mass.As we said, the net force acting on the mass must be zero; therefore,F F F . Remember, these are vectors so they must add as vectors 5 to beT1 T 2 g0zero, as shown in the following vector diagram:Figure 23: Hanging Mass Vector DiagramNote that our vector diagram starts and ends at the same point; therefore, theresultant vector (the net force) is zero.Since force is a vector, the components of the net force on a body in equilibriummust each be zero, soand 0(1.3)F x5 When drawn head to tail, they must return to the starting point.RRHS Physics Page 91

0(1.4)F yLooking at the components in the x and y direction separately, this tells us that inthe x directionF 0and in the y directionFFFx1x2xF1y1yF1x2xFF2 y2 y00 F1x2x1y2y 0 0F F F 0FFFyFgg FgThe requirement that the net force be zero (which provides translational equilibrium)is only the first condition for equilibrium. The second condition will be discussed insection 5.3.2.Equilibrant ForceIf the vector sum of all of the forces acting on an object is not zero, there will be anet force in some direction. There is a single additional force that can be applied tobalance this net force. This additional force is called the equilibrant force. Theequilibrant force is equal in magnitude to the sum of all of the forces acting on theobject, but opposite in direction.ExampleA 20.0 kg sack of potatoes is suspended by a rope. A man pushes sideways with aforce of 50.0 N and maintains this force so that the sack is in equilibrium. What is thetension in the rope and what angle does the rope make with the vertical?SolutionWhile the man is pushing sideways, the rope will be at an angle as shown in thepicture below:RRHS Physics Page 92

m 20.0kgFFpT 50.0N ?Fg mg (20.0)(9.80)196NOur free body diagram will look like this:Since the bag is in equilibrium, the net force must be zero; therefore, F F F 0 .T g pRRHS Physics Page 93

Since the force of gravity and the force of the person pushing are perpendicular, thePythagorean Theorem can be used to find the magnitude of the tension and usingtrigonometry will allow us to calculate the angle.F F F2 2T g p(196) (50.0) 202N2 2FtanFpg50.0196 14.3RRHS Physics Page 94

Check Your LearningJoe wishes to hang a sign weighing 750.0 N so that cable A attached to the storemakes a 30.0 angle as shown in the picture below. Cable B is attached to anadjoining building and is horizontal. Calculate the necessary tension in cable B.Fg 750.0N 30.0FTB ?Free Body DiagramSince the sign is in equilibrium, we know that the net force is zero andRRHS Physics Page 95

FTA Fg FTB 0Vector DiagramFtan FTBFTBtan 30.0750.0FTBg 433NRRHS Physics Page 96

1.3.1 In Class or Homework Exercise1. Find the unknown mass in the diagram below:Free Body Diagramwhere F2 225N.TSince F1 F2 F 0 ,T T gF FT 2y T 2sin 35(225)sin 35130NF cos35T 2x FT2(225)cos35180NRRHS Physics Page 97

Looking at the vector diagram, we know thatF FT1x T 2x180NsoFT2ytan 58180F 290NT2yLooking at the vertical components of the forces,F F FT 2y T1y g130 290 m(9.80)m 43kg2. A sign with a mass of 165 kg is supported by a boom and a cable, as shownin the diagram below. The cable makes an angle of 36 o with the boom. Findthe magnitude of the force exerted by the boom and the cable on the sign.Ignore the mass of the boom.m 165kg 36FFBC ? ?Fg mg (165)(9.80)1620NRRHS Physics Page 98

The purpose of the boom is to hold the sign away from the building; it musttherefore exert a force to the right.FB FC Fg 0Ftan 36F1620tan 36FFBgBB 2200NFsin 36F1620sin 36FFCgCC 2800N3. Find the tensions FT1and FT 2in the two strings indicated:RRHS Physics Page 99

F F FT1 T 2 g 0F FT1xT1 0.92Fcos 23T1F FT1yT1 0.39Fsin 23T1F FT 2x T 2 0.57Fcos55T 2F FT 2y T 2 0.82Fsin 55T 2F FT 2x T1x0.57F 0.92FFT2 T11.6FT2 T1F F FT 2y T1y g0.82F 0.39F FT 2 T1gUsing substitution, this system of equations can be solved:RRHS Physics Page 100

0.82F 0.39F mgT2 T10.82(1.6 F ) 0.39F34(9.80)T1 T11.7F 333FT1T1 200NF1.6FT2 T11.6(200) 320N4. In the diagram shown below, block A has a mass of 10.5 kg and block B has amass of 52.6 kg. The string runs from Block B to the wall. The segment ofstring from block B to point P on the string is horizontal. The friction betweenblock B and the table is unknown. Find the minimum coefficient of frictionbetween block B and the table that would prevent block B from moving.mmAB ?10.5kg 52.6kgFgA m gA10.5(9.80)103NFgB m gB (52.6)(9.80) 515NFree Body Diagram for block BRRHS Physics Page 101

Free Body Diagram for point PSince F1 F2 F 0T T gFtan 30.0FT1gAT1103tan 30.0FFT1 178NFrom the free body diagram for block B, we know thatF FFFT1T1T1178 (515) f FN FgB0.3465. Three students are pulling ropes that are attached to a car. Barney is pullingnorth with a force of 235 N; Wilma is pulling with a force of 175 N in a direction23 o E of N; Betty is pulling with 205 N east. What equilibrant force must afourth student, Fred, apply to prevent acceleration?F 235 N,N1F 175 N,23 E of N2F 205 N,E3F ?4RRHS Physics Page 102

F1 F2 F3 F4 0F F sin 232x2(175)sin 23 68NF F cos 232y2(175)cos 23160NF1 F2 y F4y 0F F F 01 2 y 4 yF4 y235 160 400NF2 x F3 F4x 0F F F 02x3 4xF4x68 205 273NF F F2 24 4x4 y273 400 480N2 2FtanF4x4 y273400 34RRHS Physics Page 103

F4 480 N,34 W of S6. Your mother asks you to hang a heavy painting. The frame has a wire acrossthe back, and you plan to hook this wire over a nail in the wall. The wire willbreak if the force pulling on it is too great, and you don't want it to break. If thewire must be fastened at the edges of the painting, should you use a shortwire or a long wire? Explain.If we look at one side of the string (since it is symmetrical)FcosFFTTyTFTycosSince FTyis the same for each wire (because the weight of the picture is thesame and it is this component that supports the weight), as the angle decreases (and cos increases) the tension in the wire decreases. Youshould therefore use a long wire.RRHS Physics Page 104

7. When lifting a barbell, which grip will exert less force on the lifter's arms: onein which the arms are extended straight upward from the body so that are atright angles to the bars, or on in which the arms are spread apart so that thebar is gripped closer to the weights? Explain.The vertical component of the force exerted by (and on) the lifter’s arms is thesame in each case, since it is this force that holds the barbell in the air. Themore spread out the lifter’s arms are, the greater the horizontal componentthat is being exerted. Since having the arms extended upward exerts thesame vertical force and less horizontal force as compared to having the armsspread out, having the arms extended upward will exert less total force on thelifter’s arms.RRHS Physics Page 105

1.3.2 Torque and Rotational EquilibriumUntil now, we have treated all objects as point sources when drawing our free bodydiagrams; in other words, we were not concerned with the where on the object theforce was being applied. In many situations, however, it makes a difference on whichpart of the object we apply a force.Even if all of the forces acting on an object balance, it is possible for the object not tobe in total equilibrium. Consider a board where equal forces are applied at oppositeends of the board, but one up and one down.Obviously, even though the forces are equal and opposite and the board as a wholeshould not move up or down, the board will begin to spin. It is not in rotationalequilibrium. Rotational equilibrium refers to the situation where there is no rotarymotion. To examine this more, we must introduce the notion of a torque.TorqueA torque has the same relationship to rotation as force does to linear movement. Itcan be thought of as a twisting force. To measure the rotating effect of a torque, it isnecessary to choose a stationary reference point for the measurements (the pivotpoint). This pivot point can be chosen arbitrarily, since the point of rotation is oftennot known until the rotation begins. 6The size of a torque depends on two things:1. The size of the force being applied (a larger force will have a greater effect)2. The distance away from the pivot point (the further away from this pivot, thegreater the effect).A line drawn from the pivot to the force that is providing the torque is known as thetorque arm. It is along this torque arm that the distance should be measured.A torque is the product of a force multiplied by a distance from the pivot.6 If there is a natural pivot point (for example, on a see-saw) then it usually makes sense to choosethis as the pivot point. However, it is not necessary to do so.RRHS Physics Page 106

Fr(1.5)where it is only the component of the force that is perpendicular to the torque armthat contributes to the torque (try opening a door by pushing parallel to the door – itdoes not work!).As can be seen from equation (1.5), the units for torque are Nm · , if the force F is innewtons and the distance r from the pivot is in metres. You may recall from Unit 4that a Nm · was defined as a joule; however, this unit is not called a joule whencalculating torques as the force and the distance are perpendicular, not parallel (asthey are when calculating work).Example 1Suppose that you are trying to open a door that is 70.0 cm wide. You are applying aforce of 68 N 10.0 cm from the outer edge of the door, but you are pushing at anangle of 75 o from the surface of the door. What torque are you applying on the door?SolutionFigure 24: Overhead view of door in Example 1As can be seen in the overhead view of the door, the force is being applied 60.0 cmfrom the natural pivot point, which is the hinge. Remember also, that it is only thecomponent of the force that is perpendicular to the torque arm (the door) thatcontributes to the torque.r 0.600mF 68N 75 ?RRHS Physics Page 107

Fr Fry Fr sin(68)(0.600)sin 75 39NmRRHS Physics Page 108

Rotational EquilibriumWhile forces were described using up, down, left, right, etc., torques are describedusing the terms clockwise and counterclockwise. A clockwise torque added to anequal (in magnitude) counterclockwise torque will be zero. Rotational equilibrium isattained if the sum of all of the torques is zero. 0(1.6)This is the second condition for static equilibrium. Just as the components of a forceare positive or negative depending on their direction, so is a torque. If a clockwisetorques is considered to be positive, then a counter-clockwise torque must beconsidered to be negative; therefore, equation (1.6) can also be expressed by sayingthat the sum of all of the clockwise torques must equal the sum of all of the counterclockwisetorques. 0cwccwcw ccw(1.7)As we have seen, there are two conditions for static equilibrium:1. The sum of the forces is zero (providing translational equilibrium), and2. The sum of the torques is zero (providing rotational equilibrium).It is important to remember that when calculating the torques, all distances must bemeasured from the pivot point. Any convenient location can be chosen for the pivotpoint.An equilibrant force must provide both translational and rotational equilibrium. Whenfinding an equilibrant force to satisfy both of these conditions, it is necessary to findboth the force itself (magnitude and direction) and the location of application.Centre of Gravity One of the forces often involved in calculating the torques on anobject is the force of gravity. Before dealing with torques, we were not usuallyconcerned with the location of the force on a body, but for calculating torques, this isimportant. Where does gravity act on a body? Of course, it acts on every particle inthe body, but there is a point called the centre of gravity (cg) where the entire force ofgravity can be considered to be acting. The center of gravity is the point at which wecould apply a single upward force to balance the object. For a mass with a uniformdistribution of mass (such as a ruler), the centre of gravity would be at the geometriccenter (the middle of the ruler).Example 2RRHS Physics Page 109

A 2.0 kg board serves as a see-saw for two children. One child has a mass of 30.0kg and sits 2.5 m from the pivot point. At what distance from the pivot must a 25.0 kgchild sit on the other side to balance the see-saw? Assume that the board is uniformand centred over the pivot.Solutionmb1212 2.0kgm 30.0kgm 25.0kgr 2.5mr ?F m gg1 1 (30.0)(9.80) 294NFg m g2 2 (25.0)(9.80) 245NDrawing a free body diagram, we haveIf we choose our pivot point here to be the centre of the board, then there is notorque due to either the force exerted by the pivot or the weight of the board, sinceboth of these forces act on the pivot point and the torque arm is zero. The onlytorques present will be due to the weight of the two children. Since we already knowboth of these forces, we do not have to consider translational equilibrium in thisproblem.cwF rg2 2 g1 1245 r (294)(2.5)r22 ccw F r 3.0mThe second child must be 3.0 m away from the pivot on the other side of the board.Example 3RRHS Physics Page 110

A 4.0 m platform with a uniform distribution of mass has a 3.2 kg box 0.80 m from theleft end. The mass of the platform is 2.0 kg. Calculate the size and location of therequired equilibrant force.SolutionThe equilibrant force is the single force that will provide equilibrium (bothtranslational and rotational). In this case, the direction of the equilibrant force isobviously upward since all of the forces on the platform are downward.Since the platform has a uniform distribution of mass, the center of gravity is at thecenter of the platform. We can guess at the approximate location of the upwardequilibrant force.In this situation, there is no natural (or fixed) pivot point – the platform could rotateabout a number of points. In this situation, it is completely arbitrary where we pick thepivot point. For this problem, we will choose the left end to be the pivot point (asindicated by the X in the diagram).mmbpbeqp 3.2kg 2.0kgr 0.80mrr 2.0m ?Fgb m gb (3.2)(9.80) 31.4NFgp m gp (2.0)(9.80)19.6NTranslational EquilibriumTo satisfy translational equilibrium, the forces must balance – in this case, the totalforce up must equal the total force down.RRHS Physics Page 111

Fy00 F F FF F F31.4 19.6 51Neq gb gpeq gb gpRotational EquilibriumTo satisfy rotational equilibrium, the torques must balance – the clockwise torquesmust equal the counter-clockwise torques. In this example, the forces of gravity onthe platform and on the box provide clockwise torques; the upward equilibrant forcewill provide a counter-clockwise torque.0 0cwccw(31.4)(0.80) (19.6)(2.0) (51)rcwccwF r F r F rgb b gp p eq eqreq 1.3meqFeq 51Nup, 1.3 m from the left endRRHS Physics Page 112

Check Your LearningA uniform 1500 kg bridge, 20.0 m long, supports a 2200 kg truck whose centre ofmass is 5.0 m from the right support column as shown in the diagram below.Calculate the force on each of the vertical support columns.This time there are two unknown forces that must be found. We will use the left endas the pivot point in this problem; since this is at the location of one of our unknownforces, this eliminates an unknown torque (since the torque arm is zero)mbt2121bt 1500kgm 2200kgF ?F ?r 10.0mr 15.0mr 20.0mr 0Fgb m gb (1500)(9.80)14700NFgt m gt (2200)(9.80) 21600NTranslational EquilibriumTo satisfy translational equilibrium, the net force must be zero:RRHS Physics Page 113

Fy0F F F F 01 2gbgtF F 14700 216001 2F F 36300N1 2Rotational EquilibriumTo satisfy rotational equilibrium, the sum of the torques must be zero:0 0cwgb bccw22 2(14700)(10.0) (21600)(15.0) F (20.0)cwF r F r F rgt t 2ccwF 23600N 24000NSince we know one force, we can now find the other one:F F3630011 2F 23600 36300F 12700N1 13000NNotice that the support closest to the truck is exerting more force, which is to beexpected.RRHS Physics Page 114

1.3.2 In Class or Homework Exercise1. A person is trying to open a door that is 90.0 cm wide. If there is a spring on adoor 5.0 cm from the hinges which exerts a force of 60.0 N to keep the doorclosed,a. How much force must be used to open the door if the force is applied atthe outer edge of the door?Using the hinge as the pivot point, we haveF 60.0Nsr 5.0cm 0.050msr 90.0cm 0.900mpFp ?The torque from the person must be greater than the torque from thespring: pF rpp pps F rs sF (0.900) (60.0)(0.050)F 3.3Nb. How much force must be used if the force from the person is applied15.0 cm from the hinges?Using the hinge as the pivot point, we haveF 60.0Nsr 5.0cm 0.050msr 15.0cm 0.150mpFp ?The torque from the person must be greater than the torque from thespring: pF rpp pps F rs sF (0.150) (60.0)(0.050)F 20. NAs can be seen, it will be harder (requires more force) if the personapplies the force closer to the hinge.RRHS Physics Page 115

2. A 60.0 kg person is sitting 1.2 m from the pivot on a see-saw. A 50.0 kgperson is sitting 0.90 m away from the pivot on the other side. Where must a22.0 kg child sit to balance the see-saw?m 60.0kg1231r 1.2mm23 50.0kgr 0.90mmr ? 22.0kg Fr1 1 1 (60.0)(9.80)(1.2) 706Nm Fr2 2 2 (50.0)(9.80)(0.90) 441NmThe 22.0 kg child must sit on the same side as the 50.0 kg person, since thisperson is exerting a smaller torque. cw1 2 3706 441Fr3ccw 3 3265 (22.0)(9.80) rr 1.2m3The 22.0 kg child must sit 1.2 m away from the pivot on the same side as the50.0 kg person.3. A long board is holding a person in the air. The person has a mass of 75.0 kgand is located 2.0 m from one end. The 10.0 m board has a mass of 10.0 kg,and its center of gravity is located 4.0 m from the same end as the person.The board is being held up by two students, one at either end. What force isrequired by each student to hold the board up?RRHS Physics Page 116

We will use the left end as the pivot point in this problem;m 75.0kgmF2121pbpb 10.0kg ?F ?r 2.0mr 4.0mr 10.0mr 0Fgp m gp (75.0)(9.80) (10.0)(9.80) 735N 98.0NFgb m gTranslational EquilibriumTo satisfy translational equilibrium, the net force must be zero:bFy0F F F F 01 2gbgpF F 98.0 7351 2F F 833N1 2Rotational EquilibriumTo satisfy rotational equilibrium, the sum of the torques must be zero:RRHS Physics Page 117

0 0cwgb bccw22 2(98.0)(4.0) (735)(2.0) F (10.0)cwF r F r F rgp p 2ccwF 186N 190NSince we know one force, we can now find the other one:F F83311 2F 186 833F 647N1 650N4. Find the equilibrant force (magnitude, direction, and location) in the followingdiagram.Using the left end as the pivot,F 200. N,downF123123eqeq 1325 N,downF 500. N,upF ?r 0r 2.0mr 3.5mr ?Translational Equilibrium (using down as positive)RRHS Physics Page 118

Fy0F F F F 01 2 3200. 1325 500. F 0Feqeqeq 1025NSo we need 1025 N upward.Rotational Equilibrium (using the left end as the pivot)0 0cwccwcw2 2 3 3ccwF r F r F r1325(2.0) (500.)(3.5) (1025)rreq 0.88meq eqF 1025 N up, 0.88 m from the left endeqeq5. Find the equilibrant force (magnitude, direction, and location) which willstabilize the following beam:Using the left end as the pivot,RRHS Physics Page 119

F 425 N,downF12341234eqeq 600. N,downF 300. N,downFF 550. N,up ?r 0r 1.5mr 3.0mr 2.5mr ?Translational Equilibrium (using down as positive)Fy0F F F F F 01 2 3 4425 600. 300. 550. F 0Feqeqeq 775NSo we need 775 N upward.Rotational Equilibrium (using the left end as the pivot)0 0cwccwcw2 2 3 3 4 4ccwF r F r F r F r600.(1.5) 300.(3.0) (550.)(2.5) (775) rreq 0.55meq eqF 775 N up, 0.55 m from the left endeqeq6. In the following diagram, determine the magnitude, direction, and point ofapplication of the necessary equilibrant force.RRHS Physics Page 120

Using the left end as the pivot,F 50. N,downF12341234eqeq 35 N,downF 100. N,30.0down from the boardFF75 N, 20.0 up from the board ?r 0r 2.3mr 4.7mr 5.9mr ?Since F3and F4are not perpendicular or parallel to the torque arm, we mustcalculate the components:F F sin 30.03y3(100.)sin 30.0 50.0NF cos30.03x F3 (100.)cos30.0 86.6NF F sin 20.04y4(75)sin 20.0 26NF F cos 20.04x4(75)cos 20.0 70. NTranslational Equilibrium (using down and to the right as positive)Fy0F F F F F 01 2 3 y 4 y eq50. 35 50.0 26 F 0Feqyeqy 109NFx0F F F 03x 4x eqx86.6 70. F 0 157NSo we need 109 N upward and 157 N to the left.RRHS Physics Page 121Feqxeqx

F eqF eq(109) (157)191N2 2109tan157 34.8Rotational Equilibrium (using the left end as the pivot)0 0cwccwcwccwF r F r F r F r2 2 3 y 3 4 y 4 eq eq35(2.3) (50.0)(4.7) (26)(5.9) (109) rreq 1.5mF 191 N,34.8 above the torque arm and to the left, 1.5 m from the left endeqeq7. A 5.0 m long ladder leans against a wall at a point 4.0 m above the ground.The ladder is uniform and has a mass of 12.0 kg. Assuming that the wall isfrictionless (but the ground is not) determine the force exerted on the ladderby the wall.Using the bottom of the ladder as the pivot,m 12.0kglwlr 2.5mrFw 5.0m ?Since the wall is frictionless, it can only exert a normal force perpendicular toitself.RRHS Physics Page 122

4.0sin5.0 53It is only the components of the force perpendicular to the ladder thatcontribute to the torque (using the ladder as the torque arm).Fgy F cosg(12.0)(9.80)cos53 70.8NUsing the bottom of the ladder as the pivot,0 0cwccwcwF rgy gccwwy w(70.8)(2.5) F (5.0)Fwy F rwy 35.4NFsin Fwyw35.4sin 53FFww 44N8. A sign with a mass of 165 kg is supported by a 35 kg boom (with a uniformdistribution of mass and a length of 1.6 m) and a cable, as shown in theRRHS Physics Page 123

diagram below. The cable makes an angle of 36 omagnitude of the force exerted by the cable.with the boom. Find theUsing the boom as the torque arm (and the left end as the pivot),mmbsTbsT 35kg 165kgr 0.8mr 1.6mrF 1.6m ?Fgb m gb (35)(9.80) 340NFgs m gs (165)(9.80)1620N0 0cwccw(340)(0.80) (1620)(1.6) F (1.6)cwccwF r F r F rgb b gs s Ty TFTy Ty1790NFsin FTyT1790sin 36FTTF 3000NRRHS Physics Page 124

9. Calculate the forces F1and F2that the supports exert on the diving boardshown below when a 50.0 kg person stands at its tip.a. ignoring the mass of the boardUsing the left end as the pivot point,m 50.0kg12p12pr 0r 1.0mr 4.0mF ?F ?Fgp m gp (50.0)(9.80) 490. NTranslational Equilibrium (using up as positive)Fy0F1 F2 FgpF1 F2 Fgp 0 0F F490.1 2Rotational Equilibrium (using the left end as the pivot)RRHS Physics Page 125

0 0cwccwcwF rgp p2 F r2 2(490.)(4.0) F (1.0)F 2ccw 1960N1F F1 21490.F 1960 490.F 1470NSo F2is applying an upward force of 1960 N and F1is applying adownward force of 1470 N. Notice that our original assumption that F1was upward (in the free body diagram) was wrong; the negative signthat we obtained indicated this.b. If the board has a mass of 40.0 kg (uniformly distributed)Using the left end as the pivot point,m 50.0kgm12bp12pbr 0 40.0kgr 1.0mr 2.0mr 4.0mF ?F ?Fgp m gp (50.0)(9.80) 490. NFgb m gb (40.0)(9.80) 392 NTranslational Equilibrium (using up as positive)RRHS Physics Page 126

Fy0F1 F2 Fgp Fgb 0F F F F 01 2gpgbF F 490. 3921 2F F 882N1 2Rotational Equilibrium (using the left end as the pivot)0 0cwgp pccw22 2(490.)(4.0) (392)(2.0) F (1.0)cwF r F r F rgb b 2ccwF 2740N1F F1 21490.F 2740 882F 1860NSo F2is applying an upward force of 2740 N and F1is applying adownward force of 1860 N. Notice that our original assumption that1Fwas upward (in the free body diagram) was again wrong; the negativesign that we obtained indicated this.RRHS Physics Page 127

1.3.3 Module SummaryIn this module you learned thato An object is said to be in static equilibrium if it is in both translationalequilibrium and rotational equilibrium.o Translational Equilibrium is achieved when the net force is zero: 0F x 0F yo Torque can be calculated using the equation Fro Rotational Equilibrium is achieved when the net torque is zero: 0RRHS Physics Page 128

Module 1.4 2D Collisions1.4.1 Conservation of MomentumYou learned in grade 11 that the total momentum of an isolated system remainsconstant. Also, if you remember from grade 11, momentum is a product of mass andvelocity ( p mv ). Since velocity is a vector, so is momentum. This vector nature ofmomentum becomes extremely important in two dimensional collisions.When you analyzed one dimensional collisions, you could show that in an isolatedsystem the momentum of each object before the collision added up to equal the totalmomentum after the collision. This still applies in two dimensional collisions, butremember that momentum is a vector so it must be added as a vector!! For acollision involving two objects in one dimension, you would writeor, sincepmv,p p p p (1.8)a b a bm v m v m v m v (1.9)a a b b a a b bwhere primed quantities mean after the collision and unprimed mean before thecollision. The vector nature of the momentum was addressed in one dimensionalsituations using positive or negative values for the velocities.In two dimensions, the vector nature of momentum does not allow simple algebraicoperations using equation (1.9). Although you can still express the conservation ofmomentum using equations (1.8) and (1.9), special attention must be paid to thevector nature of momentum. To add momentum vectors in two dimensions, avector diagram must be drawn. Equation (1.9) can only be used algebraically ifyou first break the vectors into components and then apply the equation in eachdimension.Consider the example of a ball moving to the right that collides with another ball atrest.If the collision is not head on, the two balls will go in different directions after thecollision.RRHS Physics Page 129

Just as with one dimensional collisions, the sum of all of the momentum vectors afterthe collision ( p aandcollision ( pa).p b) is equal to the total of the momentum vectors before thep p p (1.10)a a bSince momentum is a product of mass (a scalar) and velocity (a vector), themomentum vector for an object will be in the same direction as the velocity vector ofthe object; however, remember that it is momentum that is conserved, not velocity –Do not draw a velocity vector diagram when solving these problems! Themomentum vector diagram for equation (1.10) would look like this:wherecollision.ptis really justpa, since there is only one momentum vector before theThe individual momentum vectors can be found using the formula p mv . We cannow use our usual methods of component analysis for solving vector problems.If we draw our components into the momentum vector diagram, we see that themomentum is conserved in each dimension.RRHS Physics Page 130

In other words, the sum of the x components of momentum before the collision areequal to the sum of the x components of momentum after the collision.p p p a ax bx pp axbxwhere the momentum components can be found using the appropriate velocitycomponents ( p ax mavaxand p bx mbvbx).Similarly the sum of the y components of momentum before the collision are equal tothe sum of the y components of the momentum after the collision. Since the originaly momentum is zero in this example, the y momentum after the collision is still zero.0 payp 0 p p aybybyRRHS Physics Page 131

1.4.1 In Class or Homework Exercise1. A collision between two vehicles occurs at a right angled intersection. VehicleA is a car of mass 1800 kg travelling at 60. km/h north. Vehicle B is a deliverytruck of mass 3500 kg initially travelling east at 45 km/h. If the two vehiclesremain stuck together after the impact, what will be their velocity after theimpact? How much kinetic energy was lost in the collision?m 1800kgmAB 3500kgv 60. km / h N 16.7 m / s NAv 45 km / h E 12.5 m / s EvBABE ?k?pA B ti pp p p fp m vA A A1800(16.7) 30060 kgm / sp m vB B B 3500(12.5) 43750 kgm / sp Bp Ap tp p p2 2t A B(30060) (43750) 53080 kgm / s2 2p m v t t t53080 (5300) v 'v 10. m / sttpBtanpA4375030060 56RRHS Physics Page 132

vAB 10. m / s, 56E of NE m v m v1 2 1 2i 2 A A 2 B B(1800)(16.7) (3500)(12.5)1 2 122 255.2410JE1 2f 2 t t12m v (5300)(10.)52.6510J2E E Efi 2.6510 5.24105 2.610J5 52. A radioactive nucleus at rest decays into a second nucleus, an electron, and aneutrino. The electron and neutrino are emitted at right angles and havemomenta of 8.6 10 23 kg m/s and 6.210 23 kg m/s. What is the magnitudeand direction of the momentum of the recoiling nucleus?pppelectronneutrinonucleus ?238.610 /kgm s236.210 /kgm sp electronp neutrinop nucleuspi pf0 p p pelectron neutrino nucleusRRHS Physics Page 133

p neutrinop electronp nucleusp p p2 2nucleus electron neutrino (8.610 ) (6.210 )221.110 /23 2 23 2kgm spelectrontanpneutrino8.6106.210 542323pnucleus 221.1 10 kgm / s,54 from a line parallel to the neutrino3. A collision investigator is called to an accident scene where two vehiclescollided at a right-angled intersection. From skid marks, the investigatordetermined that car A, mass 1400 kg was travelling 50. km/h west beforeimpact. The two vehicles remained stuck together after impact and thevelocity of the cars after impact was 10. km/h in a direction 30.0 W of N.a. What was the mass of vehicle B?m 1400kgvvAmAABB 50. km / h W 10. km / h 30.0 W of N ?p m vA A A (1400)(50.)47.010 /kgkm hpi pp p pA B ABfSince the intersection is right angled, car B must have been travellingdirectly north.RRHS Physics Page 134

p Ap AB30.0p BpAsin p AB7.010sin 30.0 p pAB AB51.410 /4kgkm hp m vAB AB AB51.410 mAB(10.)mAB14000kgmm m mBB A AB1400 14000mB 12600kgb. How fast was vehicle B travelling before the accident?pAtan pBpB mBvB47.0105tan 30.0 1.2110 12600vBpB5 9.6 km / hp 1.2110 kgkm / hB4. Two streets intersect at a 40. angle. Car A has a mass of 1500 kg and istravelling at 50. km/h. Car B has a mass of 1250 kg and is travelling 60. km/h.If they collide and remain stuck together, what will be the velocity of thecombined mass immediately after impact?p m vA A A (1500)(50.)75000 kg km / hp m vB B B (1250)(60.)75000 kg km/ hRRHS Physics Page 135

pi pp p pA B ABfpAx pAcos 40. (75000)(cos 40. )57500 kg km / hpAy pAsin 40. (75000)(sin 40. )48200 kg km / hp p pABx B Ax75000 57500132500 kg km / hpABy pAy48200 kg km / hp p p2 2AB ABx ABy132500 482002 2141000 kg km / hp m vAB AB AB141000 (2750) vvAB 51 km / hABptanpAByABx48200132500 20.vAB 51 km / h, 20. from each car's original direction5. A billiard ball of mass 0.400 kg moving with a speed of 2.00 m/s strikes asecond ball, initially at rest, of mass 0.400 kg. The first ball is deflected off atan angle of 30.0 o with a speed of 1.20 m/s. Find the speed and direction of thesecond ball after the collision.RRHS Physics Page 136

pi pp p p f1 1 2p m v1 1 1 (0.400)(2.00)0.800 kg m / sp m v 1 1 1 (0.400)(1.20)0.480 kg m / sp cos30.01x p1 (0.480)cos30.00.416 kg m / sp psin 30.01y1 (0.480)sin 30.00.240 kg m / sp p p2x1 1x2x0.800 0.416 pp2x 0.384 kg m / sp0 p2 yp1y2y0 0.240 p2 y 0.240 kg m / sp 2 (0.384) (0.240)0.453 kg m/ s2 2RRHS Physics Page 137

p m v 2 2 20.453 0.400v22v 1.13 m / s0.240tan0.384 32.0v 21.13 m / s, 32.0 from the first ball's original path6. Billiard ball A is moving at a speed of 3.0 m/s east when it strikes an equalmass ball B at rest. The two balls are observed to move off at 45 o angles toA’s original direction – ball A goes northeast and ball B goes southeast. Whatare the speeds of the two balls after the collision?pi pfp p p A A Bp Acos 45pAmv Amvv Acos 453.0v 2.1 m / sAAp Bsin 45pAmv Bmvv Bsin 453.0v 2.1 m / sBARRHS Physics Page 138

7. A grenade of mass 10.0 kg explodes into 3 pieces in the same plane, two ofwhich A (5.0 kg) and B (2.0 kg), move off as shown. Calculate the velocity ofthe third piece C.pi pf0 p p p A B Cp m v A A A (5.0)(100)500 kg m / sp m v B B B (2.0)(200)400 kg m / sp Ax (500)cos30433 kg m / sp Ay (500)sin 30250 kg m / sp Bx (400)cos 20376 kg m / sp By (400)sin 20137 kg m / sp0 p p p Ax Bx Cx0 433 376 pCx 57 kg m / sCxp0 p p p Ay By Cy0 250 137 pCy 387 kg m/ sCyRRHS Physics Page 139

pCC 57 3872 2p 391 kg m / sp m v C C C391 (3.0) v CCv 130 m / sp Cytanp Cx38757 82vC130 m / s, 82 below the +x axisRRHS Physics Page 140

1.4.2 Elastic and Inelastic CollisionsElastic CollisionsAs you learned in grade 11, an elastic collision is one in which no kinetic energy islost; the total kinetic energy of the particles before the collision is the same as thetotal kinetic energy of the particles after the collision. For a two body collision, thiswould be expressed as1 2 1 2 1 2 1 2mava mbvb mava mbvb(1.11)2 2 2 2Remember that energy is not a vector; therefore, it is only the magnitude of thevelocity that is used in (1.11).Consider the special case where particle b is initially at rest. We now have1 2 1 2 1 2mava mava mbvb2 2 2If the mass of each particle is the same, then after cancelling the mass and the factorof one half, our conservation of energy equation (1.11) reduces tov v v(1.12)2 2 2a a bwhich is really an expression of the pythagorean theorem. Since the masses areequal, the velocity vectors are proportional to the momentum vectors. A velocityvector diagram in this situation 7 would therefore show that the vectors v aand v bwould add to give the vector va. Since the magnitudes of these vectors are relatedby the pythagorean theorem, the vector diagram must be a right angle triangle.In other words, v aand v b (andp aandp b) are perpendicular to one another; afterthis collision, the two particles move off at right angles to one another. Remember,though, that this is only true for the special case where the two objects have thesame mass, the collision is elastic, and one of the particles is initially at rest.Inelastic CollisionsAn inelastic collision is one in which the kinetic energy is not conserved; some of theenergy is transformed into other types of energy, such as thermal energy. Acompletely inelastic collision is one in which the objects stick together; some energyis lost, but a completely inelastic collision does not mean that all of the energy is lost.In this type of collision, it may be possible to calculate the amount of energy lost bycomparing the total initial kinetic energy with the total final kinetic energy.7 A velocity vector diagram can be applied here only because the masses are all the same; therefore,every velocity vector is multiplied by the same factor to obtain the corresponding momentum vectorRRHS Physics Page 141

1.4.2 In Class or Homework Exercise1. A proton travelling with speed58.2 10 m/s collides elastically with a stationaryproton. One of the protons is observed to be scattered at a 60. o angle. At whatangle will the second proton be observed, and what will be the speeds of thetwo protons after the collision?Sincea. The collision is elasticb. Both objects are the same massc. And on object was at restThe angle between the two objects after the collision must be 90 o ; the secondproton will therefore be scattered at a 30. o angle.pi pp p p f1 1 2p 1cosp1mv 1cos 60. mvv 11v 18.210554.110 /m sp 2sinp1mv 2sin 60. mvv 21v 28.210557.110 /m s2. Two cars collide at an intersection. The first car has a mass of 925 kg andwas travelling north. The second car has a mass of 1075 kg and wastravelling west. Immediately after impact, the first car had a velocity of 52.0km/h, 40.0 onorth of west, and the second car had a velocity of 40.0 km/h,50.0 o north of west. What was the speed of each car prior to the collision?RRHS Physics Page 142

m 925kgm1212121075kgv 52.0 km / h,40.0 N of Wv 40.0 km / h,50.0 N of Wv ?v ?p m v 1 1 1 925(52.0)48100 kg km / hp m v 2 2 21075(40.0)43000 kg km / hpi pp p p p 1 2 1 2fp p p1 1y2 y psin 40.0 p sin 50.01 248100sin 40.0 43000sin 50.063900 kg km / hp p p2 1x2x pcos 40.0 p cos50.01 248100cos 40.0 43000cos50.064500 kg km / hp1 1 163900 (925) vv1 m v1 69.1 km / hp2 2 264500 (1075) vv2 m v2 60.0 km / h3. Sphere A of mass 2.0 kg is travelling at 10.0 m/s west and approaches sphereB of mass 5.0 kg travelling at 5.0 m/s N. After the collision, sphere A movesaway at 4.0 m/s in a direction 35 o N of E.a. What is the final velocity of sphere B?RRHS Physics Page 143

mvAmvBAB 2.0kg10.0 m / s W 5.0kg 5.0 m / s Nv 4.0 m / s, 35 N of EAv ?Bp m vA A A (2.0)(10.0)20. kg m / sp m vB B B (5.0)(5.0)25. kg m / sp m v A A A (2.0)(4.0)8.0 kg m / spi pp p p p A B A BfpAx p cos35A8.0cos356.6 kg m / spAy p sin 35A8.0sin 354.6 kg m / sp p pA Ax Bx20. 6.6 ppBxBx 26.6 kg m / spp p pB Ay By25 4.6 pByBy 20.4 kg m/ sRRHS Physics Page 144

p ( p ) ( p )2 2B Bx By(26.6) (20.4)2 233.5 kg m / sp m v B B B33.5 (5.0) v Bv 6.7 m / sBp Bxtanp By26.620.4 53v 6.7 m / s,53 W of NBb. Was the collision elastic?E m v m v1 2 1 2ki 2 A A 2 B B(2.0)(10.0) (5.0)(5.0)1 2 122 2160JE m v m v 1 2 1 2kf 2 A A 2 B B(2.0)(4.0) (5.0)(6.7)1 2 122 2130JSince energy was lost, the collision was not elastic.4. A particle of mass m travelling with a speed v collides elastically with a targetparticle of mass 2m (initially at rest) and is scattered at 90.0 .a. At what angle does the target particle move after the collision?b. What are the particles' final speeds?c. What fraction of the initial kinetic energy is transferred to the targetparticle?5. A billiard ball is moving North at 3.00 m/s, and another is moving East with aspeed of 4.80 m/s. After the collision (assumed elastic), the second ball ismoving North. What is the final direction of the first ball, and what are theirfinal speeds?6. A billiard ball of mass ma 0.40 kg strikes a second ball, initially at rest, ofmass mb 0.60 kg. As a result of this elastic collision, ball A is deflected at anangle of 30 o and ball B at 53 o . What is the ratio of their speeds after thecollision?RRHS Physics Page 145