Pulley on Incline: Finding Forces & Tension

[Riman643]

Homework Statement

Body A in the figure weighs 93 N, and body B weighs 57 N. The coefficients of friction between A and the incline are μs = 0.40 and μk = 0.22. Angle θ is 47°. Let the
positive direction of an x axis be up the incline. What is the acceleration of A if A is initially (a) at rest, (b) moving up the incline, and (c) moving down the incline?

Relevant Equations

fs = Nμs
fk = Nμk
F = ma

a) I figured this part out. Because A is at rest that means the acceleration is 0

b&c) I am completely lost how to go about this. I drew a free body diagram for A and I was able to determine 4 forces acting on it.

1: The Gravitational force on A (G_A). Using the angle (θ), and the incline as my x-axis I was able to determine
G_Ax = G_Asin(θ) = 93sin(47) = 68.02 N
G_Ay = G_Acos(θ) = 93cos(47) = 63.43 N

2: The Normal force on A (N_A). This is equal to G_Ay.
N_A = G_Ay = 63.43 N

3: The static and kinectic frictional forces on A (f_sA, f_kA). I found these using the static and kinetic friction coefficients (μ_sA, μ_kA) and the Normal force.
f_sA = N_Aμ_sA = 0.40(63.43) = 25.37 N
f_kA = N_Aμ_kA = 0.22(63.43) = 13.95 N

4: The Tension force on A (T_A). It is very tempting to say that it is equal to weight of B, but for the forces to be at rest or for A to move up the ramp T_A would have to be much greater.

I really need help in 1) Finding T_A and 2) Figuring out how the two frictional forces play a role in this. Since A would already be moving in either direction I feel like I would need to use kinetic friction somehow, but then, what would the point of including static friction in the problem? Any help/tips would be greatly appreciated.

 

[haruspex]

Because A is at rest

It is initially at rest. It is not guaranteed to remain at rest.

fs = Nμs

That is not generally true. What is the correct version?

very tempting to say that it is equal to weight of B, but for the forces to be at rest or for A to move up the ramp TA would have to be much greater.

Don't confuse motion with acceleration.

 

[Riman643]

Thanks for the reply but I am still lost. I got part “a” of this problem right. I know the acceleration is the net force of the block in one direction divided by the mass of A. What I am having trouble determining is how to calculate the net force. Going up the ramp the net force would have to be greater than the static force and gravitational force in the x direction. But if it’s moving I would also have to factor in the kinetic friction force right? I’m just confused on how to tie it all together.

 

[haruspex]

I got part “a” of this problem right.

You got the right answer by chance; the reasoning is wrong.

First step is to decide whether static friction is enough to prevent motion. For that, you need to get your standard equation for static friction right. Can you see what is wrong with the equation you quoted? Hint: it's the bit in the middle.

For parts b and c you are given that the block is moving initially, so static friction is irrelevant.

 

[Riman643]

You got the right answer by chance; the reasoning is wrong.

Could you explain how? If the object is at rest or at a constant velocity that would mean the acceleration is zero, right?

First step is to decide whether static friction is enough to prevent motion. For that, you need to get your standard equation for static friction right. Can you see what is wrong with the equation you quoted?

Would it be that fsA <= NAμsA?

For parts b and c you are given that the block is moving initially, so static friction is irrelevant.

Okay so that would mean if the block is moving up the ramp (along the x - axis in the free body diagram) that the net force would have to be greater than the kinetic friction and the gravitational force in the x direction.
So:
T_A > f_kA + G_Ax

 

[haruspex]

If the object is at rest or at a constant velocity that would mean the acceleration is zero, right?

Yes, but you are not told it remains at rest. You are told it is initially stationary; it could immediately accelerate. A stone thrown straight up is instantaneously at rest at the top of its flight, but it accelerates down at g all the while.

Would it be that fsA <= NAμsA?

Yes.

Okay so that would mean if the block is moving up the ramp (along the x - axis in the free body diagram) that the net force would have to be greater than the kinetic friction and the gravitational force in the x direction.

No. As I posted you are confusing motion with acceleration. Consider the stone again. When you throw it up it moves up, but the acceleration is down.

 

[Riman643]

No. As I posted you are confusing motion with acceleration. Consider the stone again. When you throw it up it moves up, but the acceleration is down.

So would that mean that the acceleration when it is moving up or down the ramp would be the same? And how would I figure this out? I tried subtracting the Tension force from the sum kinetic friction force and gravitational force in the x direction and found the difference to be 2.99 N. I then divided this force by the mass to give me the acceleration but it was wrong.

 

[haruspex]

So would that mean that the acceleration when it is moving up or down the ramp would be the same?

Only if all the forces are the same in the two cases, in magnitude and direction. Would they be?

tried subtracting the Tension force from the sum kinetic friction force and gravitational force in the x direction and found the difference to be 2.99 N

You don't know the tension yet, so you can't do that.
There are two masses, undergoing the same magnitude of acceleration and subject to opposite ends of the same tension. You need a FBD for each.

 

[Riman643]

Only if all the forces are the same in the two cases, in magnitude and direction. Would they be?

Yes? I have no idea at this point.

 

[haruspex]

Yes? I have no idea at this point.

Think about the kinetic friction force direction.

 

[Riman643]

Thanks! I was able to solve it. Just had to step back and think about a little.