# Finding the acceleration of two masses on a pulley system

• Jpyhsics
Can I assume the net acceleration for m2 is the one given in the problem?Yes, assuming the conditions are met.Yes.f

## Homework Statement

Two blocks of the masses m1=7.40 kg and m2=m1/2 are connected via a massless pulley and massless string. The system is currently in equilibrium but is about to start sliding, if m2 would increase even by a bit. For the friction between the surface and m1 assume that µs=µk. The block m2 is suspended from the pulley and is free to swing. What will be the acceleration of m1 if the surface starts to move in a horizontal direction to the right with an acceleration a=5.43 m/s2? Take +x direction to be toward the left. Give your answer in m/s2.

F=ma

## The Attempt at a Solution

For m1
ΣFx=FT
m1a1=FT

For m2
ΣFy=FT-mg
m2a2=FT-mg

I am not sure how to use the given acceleration in the problem. Also, does µs=µk mean there is no friction?If anyone could help me with this, I would be grateful.

how to use the given acceleration in the problem.
The string has constant length. What does that tell you about the three accelerations? Be careful with signs.
does µs=µk mean there is no friction?I
No, it means what it says, that the two coefficients are equal. Remember you are given information about the conditions before the surface accelerates.

Which way around is the diagram, is the suspended mass on the left or the right?

The string has constant length. What does that tell you about the three accelerations? Be careful with signs.

No, it means what it says, that the two coefficients are equal. Remember you are given information about the conditions before the surface accelerates.

Which way around is the diagram, is the suspended mass on the left or the right?
This is what I have so far. Since the accelerations are the same then why do we have to calculate to find the acceleration of mass 1? Do we have to account for friction? How would I proceed based on what I have written down?

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Since the accelerations are the same
Which accelerations are the same? There are the three horizontal accelerations of the surface and the two masses, and the vertical acceleration of the suspended mass.

So are the net accelerations of the 2 masses the same? and does friction matter in this case?

are the net accelerations of the 2 masses the same?
No.
Start with the horizontal acceleration of the hanging mass. Can you say what that is? Consider the forces on it.
Next, consider that the string length is constant. What does that tell you about the relationship between the other three accelerations?

No.
Start with the horizontal acceleration of the hanging mass. Can you say what that is? Consider the forces on it.
Next, consider that the string length is constant. What does that tell you about the relationship between the other three accelerations?
So is the horizontal acceleration of the hanging mass -5.43? Is that the only place where the 5.43 applies?

is the horizontal acceleration of the hanging mass -5.43?
Why would it be that? What forces act on it?

Why would it be that? What forces act on it?
Well the force of the surface moving, force of tension upwards, and force due to gravity downwards? Am I missing something?

the force of the surface moving
How does the hanging mass know the surface is accelerating?

How does the hanging mass know the surface is accelerating?
It moves rightwards as the surface moves rightwards?

It moves rightwards as the surface moves rightwards?
If it "knew" that the surface was moving rightward, it might move rightward. How would it "know"?

If it "knew" that the surface was moving rightward, it might move rightward. How would it "know"?
So does that mean there is no horizontal acceleration for the hanging mass?

So does that mean there is no horizontal acceleration for the hanging mass?
If you draw a free body diagram, what forces on the hanging mass have non-zero horizontal components?

If you draw a free body diagram, what forces on the hanging mass have non-zero horizontal components?
Gravity and tension force. So what does this tell me in terms of the question?

Gravity and tension force. So what does this tell me in terms of the question?
So is there only 2 forces for m2?

Gravity and tension force. So what does this tell me in terms of the question?
Does gravity have a non-zero component in the horizontal direction?
So is there only 2 forces for m2?
What other forces do you think there are?

I believe there is only tension and gravity, would that be right?

I believe there is only tension and gravity, would that be right?
Yes.

Yes.
Can I assume the net acceleration for m2 is the one given in the question?

Can I assume the net acceleration for m2 is the one given in the question?
Certainly not. That is the horizontal acceleration of the surface. What have you concluded from the above about the horizontal acceleration of the suspended mass?

I have this exact same question, so I think that I understand that (tension force) x cos(theta) = m2 x a2(x) and (tension force) x sin(theta) = m2 a2(y) but we aren't given the angle. How would I go about solving that?

we aren't given the angle.
You are only asked about the immediate accelerations. There has been no time for the angle to change.

You are only asked about the immediate accelerations. There has been no time for the angle to change.
So that would mean the horizontal acceleration of m2 is 0, right?

You are only asked about the immediate accelerations. There has been no time for the angle to change.
So what is the acceleration given in the question used for?

So that would mean the horizontal acceleration of m2 is 0, right?
Yes.

So what is the acceleration given in the question used for?
If you answer the question I posed at the start of post #2 and again at the end of post #6 you will find out.

If you answer the question I posed at the start of post #2 and again at the end of post #6 you will find out.
So since the string has a constant length, the magnitude of a(m2)=a(m1), right?

Also I’m a little uncertain of what’s actually happening in the system, is it correct to assume the surface accelerating because of the tension force being experienced by the pulley?

magnitude of a(m2)=a(m1),
Remember that the surface, including the pulley, is accelerating.
is it correct to assume the surface accelerating because of the tension force being experienced by the pulley?
As I read it, the acceleration of the surface is imposed by some external force.

Remember that the surface, including the pulley, is accelerating.

As I read it, the acceleration of the surface is imposed by some external force.
So what do we do about friction?

So what do we do about friction?
We are concerned about the force of friction on m1 you would agree? The acceleration of the surface is already given and friction does not act directly on m2.

What is the magnitude of this frictional force? In what direction does it act? Do we have enough information to determine either or both?

We are concerned about the force of friction on m1 you would agree? The acceleration of the surface is already given and friction does not act directly on m2.

What is the magnitude of this frictional force? In what direction does it act? Do we have enough information to determine either or both?
SO the frictional force acts on mass 1 toward the right and it is supposed to be normal force times the coefficient of friction, but we are not given the coefficient, so would we have to find that?

SO the frictional force acts on mass 1 toward the right and it is supposed to be normal force times the coefficient of friction, but we are not given the coefficient, so would we have to find that?
So since it is in equilibrium can we assume forces in the x and y direction equal zero? So is the tension force equal to m2g?

Remember that the surface, including the pulley, is accelerating.

So where I’m at so far is:

1. I found μ=0.5 based off of the fact that the system began in equilibrium, and therefore
μm(1)g = m(2)g

2. ΣF(m2) = F(tension) - (m2)*g = (m2)(a2)

ΣF(m1)(x) = F(friction) - F(tension)
= μ(m1)*g - F(tension) = (m1)(a1x)

ΣF(m1)(y) = F(normal) - (m1)*g = 0

At this point I decided to mess around with the numbers, and found that the magnitude of
a1 = (1/2)(a2)

I don’t know if I’m doing this right because I always get stuck at this point because I don’t know how to use the acceleration of the surface and the pulley...

So where I’m at so far is:

1. I found μ=0.5 based off of the fact that the system began in equilibrium, and therefore
μm(1)g = m(2)g

2. ΣF(m2) = F(tension) - (m2)*g = (m2)(a2)

ΣF(m1)(x) = F(friction) - F(tension)
= μ(m1)*g - F(tension) = (m1)(a1x)

ΣF(m1)(y) = F(normal) - (m1)*g = 0

At this point I decided to mess around with the numbers, and found that the magnitude of
a1 = (1/2)(a2)

I don’t know if I’m doing this right because I always get stuck at this point because I don’t know how to use the acceleration of the surface and the pulley...
That is exactly where I am too!