Physics 206 Example Problems Newton's Laws of Motion

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$MATH 219 - Differential Equations Quiz 3 Name Solutions Date ...$

Problem 4. Consider a box of mass 1 0 kg being lifted upward by a person standing on the surface of the earth. At a given instant the box is being accelerated upwards with an acceleration of size 5 m/s 2 . Find the force that the person exerts on the box at this instant. You must use each step in the formal routine given below in your solution to this problem! Force Problems Routine 1 . Draw a schematic of the situation. List the given quantities and the desired unknowns. 2. Choose the object to which you will apply Newton’ s Second Law. 3. Draw a free body diagram in which the forces acting on the object are drawn coming out of the object. Include an indication of your choice of coordinate systems on the diagram. Establish notation for each of the forces acting on the object. 4. Write each force in unit vector notation ( or in component form) using the established coordinate system. Do this symbolically. 5. Apply Newton’ s Second Law in each useful coordinate direction. 6. Symbolically solve for the unknown( s) in the equations of step 5. 7. Plug in numbers ( if required) to obtain a numerical value for the unknowns and/or answer any qualitative questions related to the problem. Make sure that the answer obtained makes sense. 1 . See the figure below. We are given that the mass of the object is m = 1 0 kg and that the object is being accelerated upwards at the rate 5 m/s 2 . We are to find the force exerted by the person on the box. +y Person m 2. I”ll apply the second law to the box. 3. The free body diagram is shown above. I’ ve chosen to to call the lifting force of the person F P and the gravitational pull of the earth F G. 4. In the chosen coordinate system: 5. Applying the 2nd law in the y-direction: 6. Solving for the y-component of the lifting force: FP = Fpy yˆ FG = − m gE yˆ Fnet, y = m a y m Fpy − m gE = m a y Fpy = m ay + m gE 2 FP FG
7. Plugging in the numbers: Fpy = ( 1 0 kg) 5 m/s 2 + ( 1 0 kg 9. 8 m/s 2 = 1 48 N F P = ( 1 48 N) yˆ The answer is sensible. The person must supply a force larger than the weight of the box in order to accelerate the box upwards. Problem 5. Consider a block ( initially at rest) of mass 5 kg on a table top. Assume that the surface is frictionless. Let a pushing force of 50 N directed at an angle of 30 ◦ below the horizontal act on the block. A) Find the acceleration of the block. B) Find the ( normal) force that the table exerts on the block. Pushing Force m Table The second law gives, y x ¡ FG ¡ N ¡ Fp ¡ N ¡ = Nyˆ F ¡ G = − m gE yˆ F A) Fnet, x = m ax ⇒ Fp cos( 30 ◦ ) = m ax ⇒ ax = Fp cos( 30 ◦ ) p = Fp cos( 30 ◦ ) xˆ − Fp sin( 30 ◦ ) yˆ = m ( 50) cos( 30◦ ) 5 Fnet, y = m a y N − m gE − Fp sin( 30 ◦ ) = 0 B) N = m gE + Fp sin( 30 ◦ ) = ( 5 · 9. 8) + 50 sin( 30 ◦ ) = 74 N m/s 2 = 8. 66 m/s 2 Problem 6. A block of mass 5 kg is on a plane inclined at an angle of 30 ◦ to the horizontal. The coefficient of static friction between the block and the plane is 0. 45. The coefficient of kinetic friction between ¡ the block and the incline is 0. 3. A horizontal force FH of magnitude 20 N acts on the block as shown in the figure. Assume that the inclined plane is not free to move. A) Find the components of the gravitational force on the block in a coordinate system in which the positive x-direction is down the incline and the positive y-direction is perpendicular to the incline as shown in the figure. B) What is the size of the normal force that the incline exerts on the block? C) Find the acceleration of the block. Justify your answer! m 3 y x ¡ FH θ = 30 ◦
Page 1: Problem 1 . Physics 206 Example Pro
Page 5 and 6: Solution : There is a range of mass
Page 7 and 8: I have assumed directions for the r
Page 9: Applying the second law in the x-di

Physics 206 Example Problems Newton's Laws of Motion

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