Newton's Laws of Motion

P hysics 2 06 Example P roblems Newton’ s Laws of Motion
Problem 1 . A) What is the direction of the acceleration of an object that is slowing down while heading northward? Answer : The acceleration would be southward since the net force required to cause this acceleration would be southward. The change in velocity is directed southward. B) What is the acceleration of an object thrown straight up in the air, near the surface of the earth, at the very top of its flight? Answer : The acceleration is downwards at 9. 8 m/s 2 . The only force acting on the object at the top of its flight is the gravitational force^ so the object is in free fall. The object is changing velocity as it passes through zero velocity. C) What is the direction of the net force acting on an object that is moving in a circle with constant speed. Answer : As shown in the chapter 3 text, the acceleration of such an object is directed towards the center of the circle so the net force on the object must also be directed towards the center of the circle. D) Is it possible to round a corner with constant velocity? Explain! Answer : No, to round a corner one must change direction and so one must change velocity. E) According to Newton’ s Second Law, more massive objects tend to be harder to accelerate. Why then do all objects, regardless of mass accelerate at the same rate when in free fall? Answer: Although more massive objects are harder to accelerate, the gravitational force is larger on more massive objects. These effects cancel exactly, ( with + y up) ay = Fn et , y − m gE = = − gE m m

F) Consider a person standing on a scale in an elevator. Will an accurate scale read more than, less than, or equal to the weight of the person when the elevator is slowing down while moving downwards? Explain! Answer : In this case the person is accelerating upwards and so the upward scale force must be larger than the downward gravitational force. The scale force is larger than the weight of the person. Problem 2 . Indicate whether each of the following statements is true or false. Briefly justify your answers. A) If an object is moving there must be a nonzero net force acting on the object. Answer : False. If the object has constant velocity the net force could be zero. B) An object has the same mass when on earth and when on the moon. Answer : True. Mass is an intrinsic property. C) An object has the same weight when on earth and when on the moon. Answer: False. The weight is the size of the gravitational pull on the object. This pull is less on the moon than on the earth. D) The gravitational force between two protons is greater in size than the electrostatic force between the two protons. Answer : False. The gravitational force is much smaller. Problem 3. List the four fundamental forces in nature. Answer : Gravitational, Electromagnetic, Strong Nuclear, Weak Nuclear. 1

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! 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

Force Problems Routine

FP m Person m FG

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 FP and the gravitational pull of the earth FG . 4. In the chosen coordinate system: FP = Fpy y ˆ ˆ FG = − m gE y Fn et , y = m a y Fpy − m gE = m a y Fpy = m a y + m gE 2

5. Applying the 2nd law in the y-direction:

6. Solving for the y-component of the lifting force:

7. Plugging in the numbers: Fpy = ( 1 0 kg) 5 m/ s 2 + ( 1 0 kg ˆ FP = ( 1 48 N) y The answer is sensible. The person must supply a force larger than the weight of the box in order to accelerate the box upwards.

9. 8 m/s 2 = 1 48 N

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

y m N Fp
¡

FG The second law gives, A)

ˆ ˆ ˆ N = N y FG = − m gE y Fp = Fp cos( 30 ◦ ) x − Fp sin( 30 ◦ ) y ˆ

Fn et , x = m a x ⇒ Fp cos( 30 ◦ ) = m a x ⇒ a x =

B)

Fp cos( 30 ◦ ) ( 50) cos( 30 ◦ ) = m/s 2 = 8. 66 m/ s 2 m 5 Fn et , y = m a y N − m gE − Fp sin( 30 ◦ ) = 0 N = m gE + Fp sin( 30 ◦ ) = ( 5 · 9. 8) + 50 sin( 30 ◦ ) = 74 N

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! y x FH
¡ ¡

m

3

¡

¡

¡

¡

Table

¡

x

θ = 30 ◦

y N x

FS , K FH FG

The various forces have the forms, N = Ny ˆ FH = − FH cos( 30 ◦ ) x − FH sin( 30 ◦ ) y ˆ ˆ ◦ ◦ FG = m gE sin( 30 ) x − m gE cos( 30 ) y ˆ ˆ FS , K = FS , K x could point up or down incline ˆ ˆ ˆ ˆ ˆ A) FG = m gE sin( 30 ◦ ) x − m gE cos( 30 ◦ ) y = ( 5 ) ( 9. 8 N) sin( 30 ◦ ) x − ( 5 ) ( 9. 8 N) cos( 30 ◦ ) y FG = ( 24. 5 N) x − ( 42. 4 N) y ˆ ˆ B) Applying the 2nd law in the y direction:

Fn et , y = m a y − FH + N − m gE cos( 30 ◦ ) = 0 ◦ N = FH sin( 30 ) + m gE cos( 30 ◦ ) = 20 sin( 30 ◦ ) + 42. 4 N = 52. 4 N sin( 30 ◦ ) C) The static frictional force can supply any force in the range: − µS N − ( 0. 45) ( 52. 4 N) − 23. 58 N FS FS FS µS N ( 0. 45) ( 52. 4 N) 23. 58 N

Let’ s see if the static frictional force can keep the object at rest: Fn et , x = m a x = 0 m gE sin( 30 ◦ ) + FS − FH cos( 30 ◦ ) = 0 FS = − m gE sin( 30 ◦ ) + FH cos( 30 ◦ ) = − 24. 5 N + 20 cos( 30 ◦ ) N FS = − 7. 1 8 N This force is in the range that the static frictional force can supply, so if the object starts at rest it will remain at rest. Problem 7. Consider the figure below. Let the coefficient of static friction between the block and the incline be 0. 4. A) What range of masses can the hanging block have if the system is to be in static equilibrium? B) What is the range of tensions that the string will have if the system is in static equilibrium

m2 = 10kg

m1 θ = 30o

4

Solution : There is a range of masses because static friction provides, up to its maximum value, the force it needs to maintain equilibrium. Maximum m 1 : The maximum value of m 1 is obtained by letting static friction take on its maximum value and act down the incline. For m 1 I’ ll use a coordinate system such that up is + y. For m 2 I’ ll let + x be parallel to and down the incline and + y be in the direction of the outward normal to the incline. The free body diagrams for the masses look like: y Fron2 Fron1 m1 FG1 m2 θ N FS x FG2

In the chosen coordinate systems: Fron 1 = Ty FG 1 = − m 1 gE y ˆ ˆ ˆ ˆ ˆ ˆ Fron 2 = − Tx F G 2 = m 2 gE sinθ x − m 2 gE cosθ y N = N y FS = µ S Nx ˆ Newton’ s second law applied in the y-direction on mass 1 gives, T − m 1 gE = m 1 a 1 y = 0 ⇒ T = m 1 g E Newton’ s second law applied in the y-direction on block 2 gives, N − m 2 gE cosθ = 0 ⇒ N = m 2 gE cosθ Newton’ s second law applied in the x-direction on mass 2 gives, µ S N + m 2 gE sinθ − T = 0 Using the results of Equations 1 and 2 in Equation 3 gives, Solving for m 1 gives, µ S m 2 gE cosθ + m 2 gE sinθ − m 1 gE = 0 ( 4) ( 3) ( 2) (1)

m 1 = µ s m 2 cosθ + m 2 sinθ = { ( 0. 4) ( 1 0) cos( 30 ◦ ) + ( 1 0) sin( 30 ◦ ) } kg m 1 , m ax = 8. 46 kg

Minimum m 1 : The minimum value of m 1 is obtained by setting the static frictional force to its maximum value but letting it act up the incline so that FS = − µ S Nx . The development leading to Equation ˆ 4 still holds except that a negative sign precedes the term involving µ S . That is, m 1 = − µ s m 2 cosθ + m 2 sinθ = { − ( 0. 4) ( 1 0) cos( 30 ◦ ) + ( 1 0) sin( 30 ◦ ) } kg m 1 , m ax = 1 . 54 kg The hanging block may have any mass between 1 . 54 kg and 8. 64 kg and the system will be in static equilibrium. B) Equation ( 1 ) gives the range of tensions as 1 5N to 83N. Problem 8. The coefficient of static friction between a box of mass 1 0 kg and a truck bed is 0. 4. What is the maximum possible acceleration of the truck for which the box does not slide on the truck bed? 5

( 5)

Solution : If the box is not to slip it must have the same acceleration as the truck. The forces acting on the box are a static frictional force, the normal force and the gravitational pull of the earth on the box. The static frictional force points in a direction so as to maintain no relative motion of the box and the truck. In this case that means that the static frictional force points in the direction of the acceleration since in fact, it is the force responsible for the acceleration of the box. A free body diagram is shown below. y x

N FS m FG

In this coordinate system the forces have the form, Applying Newton’ s second law in the y-direction gives, ˆ ˆ ˆ N = N y FG = − m gE y FS = µ S N x N − m gE = 0 ⇒ N = m gE

Applying the second law in the x-direction and using the above result gives, Solving for the x-component of the acceleration gives, µ S N = m a x ⇒ µ S m gE = m a x

a x = µ S gE = 0. 4 ( 9. 8) m/ s 2 = 3. 92 m/ s 2 Problem 9. Two blocks are connected by a massless, rigid rod and placed on an inclined plane as shown in the figure below. The blocks have masses m 1 and m 2 and coefficients of kinetic friction ( with the plane) of µ K 1 and µ K 2 . A) Find a symbolic formula for the acceleration of the system. B) Find a symbolic expression for the force that the connecting rod exerts on either block. C) Show that the force from part B) is zero when µ K 1 = µ K 2 . m1 m2

θ

Solution : The free body diagrams for the two blocks looks like, y FK1 m1 θ N1 Fron1 x FG1 θ FK2 N2 x FG2 y

Fron2 m2

Note that I have chosen the usual coordinate systems for inclined planes. In this coordinate system the forces that act on block 1 can be written as, ˆ ˆ ˆ ˆ ˆ N1 = N1 y FG 1 = m 1 gE sinθ x − m 1 gE cosθ y Fron 1 = Tx FK 1 = − µ K 1 N1 x
¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡

The forces that act on mass 2 can be written as,

ˆ ˆ ˆ ˆ ˆ N2 = N2 y FG 2 = m 2 gE sinθ x − m 2 gE cosθ y Fron 1 = − Tx FK 2 = − µ K 2 N2 x 6

I have assumed directions for the rod force on the blocks. All that I know for sure is that the forces that the rod exerts on the masses are equal and opposite, so T may come out to be positive or negative. Applying the second law in the y-direction on mass 1 gives, Fn et , y = m 1 a 1 y = 0 ⇒ N1 − m 1 gE cosθ = 0 ⇒ N1 = m 1 gE cosθ Fn et , y = m 2 a 2 y = 0 ⇒ N2 − m 2 gE cosθ = 0 ⇒ N2 = m 2 gE cosθ T − µ K 1 N1 + m 1 gE sinθ = m 1 a 1 x ( 6)

Similarly the second law applied in the y-direction for mass 2 gives,

( 7)

The second law applied in the x-direction for mass 1 gives, The second law applied in the x-direction for mass 2 gives,

( 8) ( 9)

A) Note that the two blocks have the same acceleration. Using Equations 6 and 7 in Equations 8 and 9 and then adding 8 and 9 gives, Solving for the acceleration component gives, a1 x = − ( µ K 1 m 1 + µ K 2 m 2 ) gE cosθ + ( m 1 + m 2 ) gE sinθ = ( m 1 + m 2 ) a 1 x − ( µ K 1 m 1 + µ K 2 m 2 ) gE cosθ + ( m 1 + m 2 ) gE sinθ m1 + m2 ( 1 0)

− T − µ K 2 N2 + m 2 gE sinθ = m 2 a 1 x

B) Dividing Equation 8 by m 1 and Equation 9 by m 2 and then taking the difference of the results gives, T Simplifying gives, 1 1 + m1 m2 − ( µ K 1 − µ K 2 ) gE cosθ = 0 m1 m2 m1 + m2 (11)

T = ( µ K 1 − µ K 2 ) gE cosθ C) Note that if µ K 1 = µ K 2 , then T = 0.

Problem 1 0. Your car is stuck in a mud hole. You are alone but have a long, strong rope. Having studied physics you tie the rope tautly to a telephone pole and pull on it sideways as in the figure below. A) Find the forces exerted by the rope on the car when θ = 3 ◦ and you are pulling with a force of 400 N and the car does not budge. B) How strong must the rope be if it takes a force F of 600 N to move the car when θ = 4 ◦ .

θ

θ

Pole

Car
F

Solution : A) Apply the second law to the junction at which force the force F is applied. Call F1 the force of the rope attached to the car on this junction and F2 the force of the rope attached to the pole acting on this junction. In the diagram take up to + y and to the right + x. In this coordinate system I can write, F = − Fy F1 = − T1 cosθ x + + T1 sinθ y ˆ ˆ ˆ Applying the second law in the x-direction gives, Fn et , x = m a x − T1 cosθ + T2 cosθ = 0 T1 = T2 ≡ T 7 F2 = T2 cosθ x + T2 sinθ y ˆ ˆ

That is I can conclude that the tensions in the two sides of the ropes are equal. Applying the second law in the y-direction gives, Fn et , y = m a y T1 sinθ + T2 sinθ − F = 0 2 T sinθ − F = 0 F 400 T= = N = 3820 N 2 sinθ 2 sin( 3 ◦ ) The size of the force that the rope pull on the car with is the tension in the rope which has just been found to be 3820 N. B) To just move the car the acceleration has to be barely more that zero. So the net force is essentially zero still. The analysis of part A) still holds so, T= 600 F = N = 4300 N 2 sinθ 2 sin( 4 ◦ )

Problem 1 1 . What size downward force F must be applied to lift the cart of weight 2000 N using the 4pulley apparatus shown below?
Ceiling

m

F

Solution . The tension in the rope going over the pulleys is constant throughout. Apply Newton’ s second law to each of the hanging pulleys. The upward force on each of these pulleys is 2 F. Since these pulleys do not accelerate ( and if the pulleys are nearly massless. . . ) the downward force on these pulleys must have size 2 F. This means that the tensions in each of the strings attached to the mass is 2 F. Therefore the net upwards force on the mass is 4 F. This force supports the weight of the mass so, W 2000 N = = 500 N 4 4 To lift the mass a force of 1 / 4 the weight of the block must be applied. The pulley arrangement is essentially a force multiplier. 4F= W⇒F= Problem 1 2 . A mass m hangs at the end of a massless string attached to the roof of a truck. The motion is such that the string makes a constant angle θ with the vertical. What must be the acceleration of the truck for this situation to be possible. ( This forms the basis for a simple means to measure the acceleration of an object. ) Solution: The forces that act on the mass are the tension in the string and the force due to the gravitational pull of the earth. With up as + y and the direction of the acceleration as + x, these forces can be written as, Frop e = T sinθ x + T cosθy FG = − m gE y ˆ ˆ ˆ Applying Newton’ s second law in the y-direction gives, Fn et , y = m a y = 0 T cosθ − m gE = 0 m gE T= cosθ 8

Applying the second law in the x-direction gives, T sinθ = m a x T sinθ ( m gE / cosθ) sinθ = = gE tanθ m m So, measuring the angle θ allows for a determination of the acceleration of the vehicle. ax =

9