Find the acceleration of a block on an inclined plane with a pulley

[isukatphysics69]

1. Homework Statement
picture

Homework Equations

f=ma

The Attempt at a Solution

Below is my attempt for part B, i am not sure what i am doing wrong here but my answer is off for acceleration

 

[TSny]

In the equation for the acceleration at the bottom of the first page, what direction are you taking to be positive acceleration?
In the first equation at the top of the second page, what direction are you taking to be positive acceleration?

In the second equation at the top of the second page, check the calculation of the numerator in parentheses.

Why wouldn't the acceleration at the bottom of the first page give you the answer to part (b)?

 

[isukatphysics69]

hi, i have the positive direction as the up the inclined plane and negative as down the inclined plane, i will recheck right now

 

[TSny]

hi, i have the positive direction as the up the inclined plane and negative as down the inclined plane, i will recheck right now

OK, I think you have the signs right. I believe the only mistake is in calculating the numerator in parentheses on the right side of the second equation on the second page.

But, is it necessary to have the second page at all? How does the acceleration as given on the bottom of the first page relate to the acceleration asked for in the problem?

Note how it is kind of awkward to have to refer to your work by line number and page number. That's why the Forum asks posters to type in their work rather than post pictures of their work, if possible. Thanks.
https://www.physicsforums.com/threads/guidelines-for-students-and-helpers.686781/ [item 5]

 

[isukatphysics69]

OK, I think you have the signs right. I believe the only mistake is in calculating the numerator in parentheses on the right side of the second equation on the second page.

But, is it necessary to have the second page at all? How does the acceleration as given on the bottom of the first page relate to the acceleration asked for in the problem?

Note how it is kind of awkward to have to refer to your work by line number and page number. That's why the Forum asks posters to type in their work rather than post pictures of their work, if possible. Thanks.

i will type work in from now on, i didn't know that, sorry. and i got the correct answer now, i guess i was plugging wrong values in calculator..might have been switching up the boxes. Thank you. Also, generally it is safe to make an assumption that the box is sliding one way or the other correct? if your acceleration is the opposite of the sign that you were expecting, then it is sliding the opposite way?

 

[TSny]

Also, generally it is safe to make an assumption that the box is sliding one way or the other correct? if your acceleration is the opposite of the sign that you were expecting, then it is sliding the opposite way?

Yes, but if the block turns out to be moving in the opposite direction to what you assumed, that could mean that the direction that you assumed for the friction force could be wrong. In that case, you would need to rework the problem with the direction of the friction force switched. You would then generally get a different magnitude of acceleration than when you had the wrong direction.

[EDIT: It is important to note that the sign of the acceleration is not directly related to the direction of motion. Suppose you take upward along the incline to be positive direction for acceleration of block A. A negative acceleration for block A doesn't mean that the block must be moving down the incline. It could be moving up the incline but slowing down, or it could be moving down the incline and speeding up. In part (b) you are told that the block is moving up the incline. You then found a negative acceleration. So it is moving up the incline and slowing down.]

 

[isukatphysics69]

Yes, but if the block turns out to be moving in the opposite direction to what you assumed, that could mean that the direction that you assumed for the friction force could be wrong. In that case, you would need to rework the problem with the direction of the friction force switched. You would then generally get a different magnitude of acceleration than when you had the wrong direction.

ok great thank you

 

[isukatphysics69]

ok i got part B and C now for part A I would just have to use static friction and set the x direction net force equal to 0 right?

doing the above didn't really help. how would i prove that acceleration is 0 when at rest, is this just a trick question or something, obviously at rest means a = 0

 

[TSny]

ok i got part B and C now for part A I would just have to use static friction and set the x direction net force equal to 0 right?

doing the above didn't really help. how would i prove that acceleration is 0 when at rest, is this just a trick question or something, obviously at rest means a = 0

In part (a) they say that block A is initially at rest. But it might not stay at rest. If it stays at rest, then of course the acceleration of the block would be zero. But if it doesn't stay at rest, the acceleration of the block will not be zero. It's like dropping a ball from rest. Just after releasing it, the acceleration of the ball is not zero.

It's important to keep in mind that an object can be at rest at some instant and yet have a nonzero acceleration at that instant.

In your written work for part (a) it looks like you are assuming that you will have a static friction force and that it will be given by $\mu_s$ times the normal force $N = mg \cos \theta$. But that isn't necessarily true. $\mu_s N$ represents the maximum possible static friction force that could occur. But you do not know if the static friction force will have its maximum value in this problem. My hint for this part would be to first find what friction force would be necessary to keep block A from sliding.

 

[isukatphysics69]

In part (a) they say that block A is initially at rest. But it might not stay at rest. If it stays at rest, then of course the acceleration of the block would be zero. But if it doesn't stay at rest, the acceleration of the block will not be zero. It's like dropping a ball from rest. Just after releasing it, the acceleration of the ball is not zero.

It's important to keep in mind that an object can be at rest at some instant and yet have a nonzero acceleration at that instant.

In your written work for part (a) it looks like you are assuming that you will have a static friction force and that it will be given by $\mu_s$ times the normal force $N = mg \cos \theta$. But that isn't necessarily true. $\mu_s N$ represents the maximum possible static friction force that could occur. But you do not know if the static friction force will have its maximum value in this problem. My hint for this part would be to first find what friction force would be necessary to keep block A from sliding.

ok so i tried something.

netforceAX = T + forceFriction - MGsin(theta)
Ma - T +MGsin(theta) = forceFriction

netforceB = T - mg
ma = T - mg
ma+mg = T
Ma - (ma+mg) +MGsin(theta) = forceFriction

But i still do not know the acceleration since it is nonzero so I'm not sure how to solve for friction

 

[TSny]

ok so i tried something.

netforceAX = T + forceFriction - MGsin(theta)

When you are typing in your equations, you can use the toolbar for formatting. Greek letters and other math symbols can be found by clicking on the Σ symbol.

So, you can make your equation more readable by typing something like

F_{net A, x} = T + f - Mgsinθ

Note that you are assuming the friction force is acting on block A in the same direction as the tension force. I'm not sure how you decided that. (You might be right.) One way I like to approach it is to first image there is no friction in the problem and decide which way the system would start to move if there were no friction. That way I can tell which way the friction force will act when friction is present.

Ma - T +MGsin(theta) = forceFriction

In this equation you are taking positive direction to be upward along the incline. So, the acceleration "a" is positive when block A accelerates upward along the incline.

netforceB = T - mg
ma = T - mg

Now "a" represents the acceleration of block B and you are taking positive direction as vertically upward. But this conflicts with your sign convention in the equation for block A. If block A accelerates upward along the slope, then block B would accelerate vertically downward. So, if you want the same symbol "a" to represent the acceleration of A and also the acceleration of B, you will need to take the positive direction for B to be vertically downward if you take the positive direction for A to be upward along the slope.

But i still do not know the acceleration since it is nonzero so I'm not sure how to solve for friction

At the beginning of the problem, you do not know if the system will start moving or not when it is released from rest. One approach is to assume the system does not move. (So, what would be the acceleration under this assumption?) How much would the static friction force need to be to keep the system at rest? Is this much static friction available?

 

[isukatphysics69]

When you are typing in your equations, you can use the toolbar for formatting. Greek letters and other math symbols can be found by clicking on the Σ symbol.
View attachment 223620

So, you can make your equation more readable by typing something like

F_{net A, x} = T + f - Mgsinθ

Note that you are assuming the friction force is acting on block A in the same direction as the tension force. I'm not sure how you decided that. (You might be right.) One way I like to approach it is to first image there is no friction in the problem and decide which way the system would start to move if there were no friction. That way I can tell which way the friction force will act when friction is present.In this equation you are taking positive direction to be upward along the incline. So, the acceleration "a" is positive when block A accelerates upward along the incline.

Now "a" represents the acceleration of block B and you are taking positive direction as vertically upward. But this conflicts with your sign convention in the equation for block A. If block A accelerates upward along the slope, then block B would accelerate vertically downward. So, if you want the same symbol "a" to represent the acceleration of A and also the acceleration of B, you will need to take the positive direction for B to be vertically downward if you take the positive direction for A to be upward along the slope.

At the beginning of the problem, you do not know if the system will start moving or not when it is released from rest. One approach is to assume the system does not move. (So, what would be the acceleration under this assumption?) How much would the static friction force need to be to keep the system at rest? Is this much static friction available?

Ohhh wait a second i think i understand i will try part a again

 

[isukatphysics69]

When you are typing in your equations, you can use the toolbar for formatting. Greek letters and other math symbols can be found by clicking on the Σ symbol.
View attachment 223620

So, you can make your equation more readable by typing something like

F_{net A, x} = T + f - Mgsinθ

Note that you are assuming the friction force is acting on block A in the same direction as the tension force. I'm not sure how you decided that. (You might be right.) One way I like to approach it is to first image there is no friction in the problem and decide which way the system would start to move if there were no friction. That way I can tell which way the friction force will act when friction is present.In this equation you are taking positive direction to be upward along the incline. So, the acceleration "a" is positive when block A accelerates upward along the incline.

Now "a" represents the acceleration of block B and you are taking positive direction as vertically upward. But this conflicts with your sign convention in the equation for block A. If block A accelerates upward along the slope, then block B would accelerate vertically downward. So, if you want the same symbol "a" to represent the acceleration of A and also the acceleration of B, you will need to take the positive direction for B to be vertically downward if you take the positive direction for A to be upward along the slope.

At the beginning of the problem, you do not know if the system will start moving or not when it is released from rest. One approach is to assume the system does not move. (So, what would be the acceleration under this assumption?) How much would the static friction force need to be to keep the system at rest? Is this much static friction available?

ok i am getting that the frictional force at rest is 33.56 Newtons which is less than the maximum static frictional force of 43.75 Newtons

This means that the acceleration is 0

 

[isukatphysics69]

i don't know how you did that subscript thing i will use some of the tools tho to make it clearer

-netForceAx = FrictionForce -MGsinθ +T
Ma = -FrictionForce + MGsinθ - T
→ (netForceB = T - mg) → (ma + mg = T)
Ma = -FrictionForce + MGsinθ - (ma + mg)
Ma = -FrictionForce + MGsinθ - ma - mg
0 = -FrictionForce + MGsinθ - 0 - mg
FrictionForce = MGsinθ - mg
FrictionForce = 102sin(40) - 32
FrictionForce = 33.56

StaticFrictionMax = .56MGcosθ = .56*102*cos(40) = 43.76

 

[TSny]

ok i am getting that the frictional force at rest is 33.56 Newtons which is less than the maximum static frictional force of 43.75 Newtons

This means that the acceleration is 0

Yes. Good.

 

[TSny]

i don't know how you did that subscript thing i will use some of the tools tho to make it clearer

The subscript and superscript tools are here

-netForceAx = FrictionForce -MGsinθ +T
Ma = -FrictionForce + MGsinθ - T
→ (netForceB = T - mg) → (ma + mg = T)

As pointed out earlier, you have a sign error here. If you take "a" to be the acceleration of block A up the incline, then the acceleration of block B will be "a" in the vertically downward direction. So, in setting up the net force on B, you should take downward forces as positive: netForceB = -T + mg. Since you are setting the acceleration equal to zero in part (a), this sign error does not affect the answer.

In your handwritten work, you had

That negative sign on the left side was correct and you got the right answers for (b) and (c).

 

[isukatphysics69]

Yes. Good.

Awsome, thank you for conformation