Chapter 5: Motion along a line

Chapter 5: Motion along a line 

Strategy for solving dynamics problems 

Model 

make simplifying assumptions (e.g., represent object as a particle) 

Visualise 

pictorial representation: draw a sketch to illustrate the motion, establish a 

coordinate system, define all variables and symbols, state what the 

problem is asking for. 

physical representation: 

- a motion diagram to establish direction of v and/or a 

- a free body diagram (FBD) to establish forces 

Solve 

mathematical representation: F net = F i = ma 

Assess 

correct units? reasonable numbers? answers the question? 

Σ i 

SMU PHYS1100, fall 2008, Prof. Clarke 1


Newton’s 2 nd law must be treated as a vector equation! 

F Σ net = Fi = ma 

i 

is really 3 equations, one for each component. 

F Σ net,x = Fi,x = max i 

F Σ net,y = Fi,y = may i 

F Σ net,z = Fi,z = maz i 

and we must solve each separately. 

In this chapter, problems are in 2-D; z-components are zero. 



Definition: An object is in equilibrium when F net = 0, and thus 

when a = 0. 

If v = 0 too, equilibrium is static. 

example: If A exerts a force of 100 N west and B exerts a force 

of 120 N south, what force must C exert to maintain the knot 

in static equilibrium? 

A 

B 

C 



Newton’s 2 nd Law: 

broken up into components: 

F Σ net = Fi = T1 + T2 + T3 = ma = 0 

i 

x: F Σ net,x = Fi,x = −T1 + T3 cosθθθθ = 0 

i 

� T 3 cosθθθθ = T 1 

y: F Σ net,y = Fi,y = −T2 + T3 sinθθθθ = 0 

i 

� T 3 sinθθθθ = T 2 

divide (2) by (1) to get: 

tanθθθθ = T 2 /T 1 

SMU PHYS1100, fall 2008, Prof. Clarke 4 

(1) 

(2) 

(1) 2 + (2) 2 � 

T3 cos2θθθθ + T3 sin2θθθθ = T3 (cos2θθθθ + sin2 2 2 

2 

θ θ θ θ ) 

2 2 2 

= T3 = T1 + T2 T 1 

y 

T 2 

θθθθ 

T 3 

T 3 cosθθθθ 

T 3 sinθθθθ 

x




F Σ net = Fi = T1 + T2 + T3 = ma = 0 

i 


i 



i 





(1) 

(2) 

(1) 2 + (2) 2 � 


2 

θ θ θ θ ) 

2 2 2 

= T3 = T1 + T2 T 1 

y 

T 2 

θθθθ 

T 3 



x




F Σ net = Fi = T1 + T2 + T3 = ma = 0 

i 


i 



i 





(1) 

(2) 

(1) 2 + (2) 2 � 


2 

θ θ θ θ ) 

2 2 2 

= T3 = T1 + T2 T 1 

y 

T 2 

θθθθ 

T 3 



x




F Σ net = Fi = T1 + T2 + T3 = ma = 0 

i 


i 



i 





(1) 

(2) 

(1) 2 + (2) 2 � 


2 

θ θ θ θ ) 

2 2 2 

= T3 = T1 + T2 T 1 

y 

T 2 

θθθθ 

T 3 



x




F Σ net = Fi = T1 + T2 + T3 = ma = 0 

i 


i 



i 





(1) 

(2) 

(1) 2 + (2) 2 � 


2 

θ θ θ θ ) 

2 2 2 

= T3 = T1 + T2 T 1 

y 

T 2 

θθθθ 

T 3 



x




F Σ net = Fi = T1 + T2 + T3 = ma = 0 

i 


i 



i 



tanθθθθ = T 2 /T 1 = 1.2 � θθθθ = 50 o 

ended here, 25/9/08 


(1) 

(2) 

(1) 2 + (2) 2 � 


2 

θ θ θ θ ) 

2 2 2 

= T3 = T1 + T2 = 24,400 � T3 = 156 N 

T 1 

y 

T 2 

θθθθ 

T 3 



x


If v = 0 (but constant since a = 0), equilibrium is dynamic. 

example: A car of weight 15,000 N is towed up a hill with an 

incline of 20 o at constant velocity. Friction is negligible. The 

tow rope is rated at 6,000 N maximum tension. Will it break? 

Notice we have chosen the x-axis to point up the hill, 

and not horizontally. See rules of thumb on page 13. 



Newton’s 2 F Σ net = Fi i 

= T + n + w = ma = 0 

x: F Σ net,x = Fi,x = T −−−− w sinθθθθ = 0 

i 

nd Law: 


y 

� T = w sinθθθθ (1) 

y: F Σ net,y = Fi,y = n −−−− w cosθθθθ = 0 

i 

� n = w cosθθθθ (2) 

Since we are only asked about T, (2) 

is of no further interest to us. From 

(1), we have: 

T = (15,000) sin(20 o ) = 5,130 N 

Since T < 6,000 N, the rope does not break. 


w 

n 

θθθθ 

T 

w cosθθθθ 

θθθθ 

w sinθθθθ 

x


Continuing with the same example, now suppose the tower 

grows impatient, and accelerates up the hill at 1.0 ms-2 . 

Whither the rope now? 

To the FBD, we add an acceleration 

vector. This changes the x-component 

of Newton’s 2 nd Law: 

x: F Σ net,x = Fi,x = T −−−− w sinθθθθ = ma 

i 

� T = w sinθθθθ + mg = w sinθθθθ + a a 

g ( g) 

� T = (15,000) sin(20 o ) + 1.0 

( ) 9.8 

= 6,660 N, and the rope breaks. 


y 

w 

n 

θθθθ 

T 

w cosθθθθ 

a 

θθθθ 

w sinθθθθ 

x


Why did we choose an inclined coordinate system in the 

previous example? 

Rule of thumb: Choose your coordinate system so that a lies 

along either the x-axis or the y-axis. 

- simplifies equations by having a appear in only one. 

Subordinate rule of thumb: If a = 0, choose your coordinate 

system so that most of the forces lie along either the x-axis or 

the y-axis. 

- reduces the task of resolving force vectors into x- and ycomponents. 



Clicker question 5.1 

Resolve T into its x- and ycomponents 

for the coordinate 

system shown. 

a) T x = T; T y = 0 

b) T x = 0; T y = T 

c) T x = T cosθθθθ; T y = T sinθθθθ 

d) T x = T sinθθθθ; T y = T cosθθθθ 


y 

θθθθ 

T 

x



Resolve T into its x- and ycomponents 


system shown. 

a) T x = T; T y = 0 

b) T x = 0; T y = T 

c) T x = T cosθθθθ; T y = T sinθθθθ 

d) T x = T sinθθθθ; T y = T cosθθθθ 


y 

θθθθ 

T x 

The x-component is adjacent 

to θθθθ and gets the cosθθθθ. 

The y-component is opposite 

to θθθθ and gets the sinθθθθ. 

T 

T y 

x



Resolve w into its x- and ycomponents 


system shown. 

a) w x = w; w y = 0 

b) w x = 0; w y = −w 

c) w x = −w cosθθθθ; w y = −w sinθθθθ 

d) w x = −w sinθθθθ; w y = −w cosθθθθ 

e) w x = w cosθθθθ; w y = −w sinθθθθ 

f) w x = w sinθθθθ; w y = −w cosθθθθ 


y 

θθθθ 

w 

x





system shown. 

a) w x = w; w y = 0 

b) w x = 0; w y = −w 





Here, we had to see that θθθθ 

is also the angle between 

w and the y-axis, and that 

both components point in 

their negative directions. 


y 

θθθθ 

w 

θθθθ 

w y 

w x 

x





system shown. 

a) w x = w; w y = 0 

b) w x = 0; w y = −w 






θθθθ 

w 

y 

x





system shown. 

a) w x = w; w y = 0 

b) w x = 0; w y = −w 






θθθθ 

w y 

θθθθ 

w x 

It doesn’t matter whether we 

measure θθθθ between the x-axis 

and the horizontal, or 

between the y-axis and the 

vertical; it’s the same angle. 

Here, the x-component points 

in the +x direction. 

θθθθ 

w 

y 

x


A model for friction… 

Any two solid surfaces in contact will “stick” to each other, even if very 

weakly, via electrostatic bonds between molecules that come in very close 

proximity to each other. When the surfaces are at relative rest, these bonds 

have more time to set up than when the 

surfaces are in relative motion. 

Thus, the maximum static friction, f s,max , 

ought to be greater than the kinetic 

friction, f k . 

SMU PHYS1100, fall 2008, Prof. Clarke (figures from Doug Davis @ Eastern Illinois University) 20


Even though the detailed physics of two surfaces in contact is very 

complex, experiments reveal a very simple model for friction. 

Consider a block at rest on the surface of a table. Gradually increase the 

horizontal force, F pull , on the cord. When F pull reaches a critical level, f s,max , 

the block suddenly moves. The frictional force, f, reduces to f k and remains 

constant, regardless of how much higher F pull gets. 

Experimentally, one finds f s,max = µµµµ s n and f k = µµµµ k n where µµµµ s > µµµµ k are the 

coefficients of static and kinetic friction respectively (Table 5.1, pg. 133). 

n 

F pull 

F pull 


f 

f s,max 

f k 

0 static region kinetic region


Note that f s is not always µµµµ s n. When F pull = 0, f s = 0 too. 

f s,max= µµµµ sn is the maximum static 

frictional force the surface can exert 

before the “sticking” molecular bonds 

break, and after which the kinetic 

frictional force takes over. 

The direction of f is tangential to the 

surface, and thus perpendicular to n. 

The direction of f s is opposite to the 

motion it is preventing. 

The direction of f k is opposite to the 

velocity. 

velocity frog would have if 

there were no f s 



example for the board (e.g. 5.6 from text): A 50 kg box is 

pushed across the floor (µµµµ k = 0.20) at a steady speed of 2.0 ms -1 . 

a) How much force is exerted on the box? 

b) If the force is stopped, how far will the box slide along the 

floor before coming to rest? 

notes 5.1 



example for the board (e.g. 5.7 from text): A 50.0 kg steel filing 

cabinet is in the back of a dump truck. The steel bed of the 

truck (µµµµ s = 0.80) is slowly raised. 

a) What is the magnitude and direction of the static frictional 

force when the bed is tilted by 20 o ? 

b) At what angle does the cabinet begin to slide? 

notes 5.2 




Consider the static or kinetic frictional force, if any, between the box and 

table when the table is tilted as shown in the four cases. The box is 

stationary in cases 1, 2, and 3 with the box just on the verge of slipping in 

case 3. The box slides with constant speed in case 4. 

Which of the following is true? 

a) 0 < f s,1 < f s,2 < f k,4 < f s,3 = µµµµ s n b) 0 = f s,1 < f s,2 < f k,4 < f s,3 = µµµµ s n 

c) µµµµ k n = f k,4 < f s,1 = f s,2 = f s,3 = µµµµ s n d) 0 = f s,1 < f k,4 < f s,2 = f s,3 = µµµµ s n 

f s,1 

f s,2 

1 2 

3 

4 


f s,3 

f k,4 

v



Consider the static or kinetic frictional force, if any, between the box and 

table when the table is tilted as shown in the four cases. The box is 

stationary in cases 1, 2, and 3 with the box just on the verge of slipping in 

case 3. The box slides with constant speed in case 4. 

Which of the following is true? 

a) 0 < f s,1 < f s,2 < f k,4 < f s,3 = µµµµ s n b) 0 = f s,1 < f s,2 < f k,4 < f s,3 = µµµµ s n 

c) µµµµ k n = f k,4 < f s,1 = f s,2 = f s,3 = µµµµ s n d) 0 = f s,1 < f k,4 < f s,2 = f s,3 = µµµµ s n 

f s,1 

f s,2 

1 2 

3 

4 


f s,3 

f k,4 

v



You push lightly against the computer sitting on your desk 

and it doesn’t move. You push harder still, and it still won’t 

budge. This is because… 

a) The friction between the computer and the table 

increases the harder you push. 

b) The amount of friction is constant, but you haven’t 

pushed hard enough to overcome this force. 

c) The friction between the computer and the table 

decreases the more you push. 

d) None of the above. 




You push lightly against the computer sitting on your desk 

and it doesn’t move. You push harder still, and it still won’t 

budge. This is because… 

a) The friction between the computer and the table 

increases the harder you push. 

b) The amount of friction is constant, but you haven’t 

pushed hard enough to overcome this force. 

c) The friction between the computer and the table 

decreases the more you push. 

d) None of the above. 



Mass, weight, and apparent weight 

mass, m, is a measure of the quantity of matter (kg). 

weight, w, is the gravitational force exerted on m (N). w = mg 

Take the earth away and your weight, w, disappears, but the amount of 

matter in your body, m, remains the same. 

apparent weight, w app, is the magnitude of the contact force 

that supports the weight, w. 

e.g., when you stand on a scale, the scale actually measures the normal 

force preventing you from falling through the scale. The normal force that 

registers on the dial is your apparent weight. 

Often, w app = w. However, if m is accelerating in a direction parallel to g, 

w app will not be the same as w. 



e.g., a man stands on a (spring) scale in an elevator 

accelerating upward with acceleration a. Find his 

apparent weight, F sp. 

F Σ net = Fi,y = Fsp − w = ma 

i 

a 

� Fsp = w + ma = w + mg g 

� Fsp = w 1 + a ( g) 

exercise: show that if the elevator is 

accelerating downward instead, the 

man’s apparent weight is given by: 

Fsp = w ( 1 −−−− 

a) g 

ended here, 30/9/08 



Weightlessness is when the apparent weight is zero, not the 

weight itself. 

e.g., what is the weight and apparent weight 

of a rock in free-fall? 

w = mg = 0 is the only force acting on m, 

(ignoring drag) and causes the rock to 

accelerate downward with acceleration g. 

Thus: 

wapp = w ( 

g 

1 −−−− g) 

= 0 

So, when m is in freefall, w app = 0 and we say m is weightless 

even though its actual weight is not zero. The term weightless 

is evidently a bit of a misnomer. 


m 

mg 

g


Air drag. Like friction, air drag is a 

very complicated process. However, 

experiments show that a relatively 

simple expression for the air drag 

force, D, usually suffices: 

D = bAv 2 

where b is the drag coefficient = 0.25 

kg s -1 for air, A is the “presentation 

cross section area” of the object, and v 

is its speed relative to the air. 

The concept of “presentation cross 

section area” is illustrated to the right. 



Terminal speed. Because air drag increases with velocity 

(in fact, as v 2 ), a falling object will reach a speed, v term , such that 

D = mg. Here, the net force on m is zero, 

a = 0, and m falls at the constant and socalled 

terminal speed, v term, given by: 

D = bAv 2 

term= mg � vterm= with b set to ¼ for air. 

4mg 

A 

e.g., for a skydiver of mass m = 75 kg 

“laying flat”, A = 1.8 m x 0.4 m = 0.72 m 2 , 

� v term = 64 ms −1 (230 km/hr) 

e.g., skydiver “standing”, A = 0.3 m x 0.4 m 

w = mg 

= 0.12 m 


2 � vterm = 157 ms−1 (560 km/hr) 

m 

D 

F net = 0 

v term


example for the board (Fig. 5.21 

from text) 

A ball of mass m and radius r 

moves up and down vertically. 

In the absence of air drag, the 

ball would be in free-fall with 

acceleration –g. 

Including air drag, 

a) what is the acceleration of 

the ball on the way up? 

b) what is the acceleration of 

the ball on the way down? 

notes 5.3 



example for the board (e.g., 

5.10 from text) 

A wooden sled with mass (including 

rider and supplies) 200 kg is pulled 

by a dog team over snow (µµµµ k = 0.06) 

with two ropes (inclined at θθθθ = 10 o to 

the horizontal) attaching the sled to 

the team. It takes the dogs 15 metres 

to reach their cruising speed of 5 ms -1 . 

a) What is the tension in the ropes while the dogs 

are accelerating (assumed to be constant)? 

b) What is the tension in the ropes at cruising 

speed? 

notes 5.4 



example for the board (e.g., 5.11 from text) 

A 100 kg box rests on the floor of a flatbed truck. The coefficients of 

friction between the box and the floor are µµµµ s = 0.3 and µµµµ k = 0.2. 

a) What is the maximum acceleration the truck can have without causing 

the box to slip? 

notes 5.5 

b) Suppose the truck is accelerating at this maximum acceleration and a 

bump causes the box to start slipping. If the distance between the centre 

of the box and the end of the flatbed is 3.0 m, how long does it take for 

the box to fall off the end of the truck? 




An elevator is going UP at constant speed. The only two forces 

on the elevator are tension, T, from a cable pulling UP, and the 

weight of the elevator and its contents, w, pulling DOWN. How 

does the magnitude of T compare to the magnitude of w? 

a) T > w 

b) T = w 

c) T < w 

d) not enough information to tell 




An elevator is going UP at constant speed. The only two forces 

on the elevator are tension, T, from a cable pulling UP, and the 

weight of the elevator and its contents, w, pulling DOWN. How 

does the magnitude of T compare to the magnitude of w? 

a) T > w 

b) T = w 

c) T < w 

d) not enough information to tell 




Consider the tension in the cable holding up an elevator 

under the following three situations: 

1) the elevator is stationary; 

2) the elevator is being accelerated upwards at 2.0 ms −2 ; 

3) the elevator is being accelerated downward at 2.0 ms −2 ; 

Rank these situations from greatest tension to lowest tension 

in the cable: 

a) 2,1,3 b) 2,3,1 c) 3,1,2 d) all the same 




Consider the tension in the cable holding up an elevator 



2) the elevator is being accelerated upwards at 2.0 ms −2 ; 

3) the elevator is being accelerated downward at 2.0 ms −2 ; 


in the cable: 





Now consider the tension in the cable holding up an elevator 



2) the elevator is moving upwards at constant speed; 

3) the elevator is moving downward at constant speed; 


in the cable: 





Now consider the tension in the cable holding up an elevator 



2) the elevator is moving upwards at constant speed; 

3) the elevator is moving downward at constant speed; 


in the cable:

Chapter 5: Motion along a line

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