3 pulley problem with attached masses

[patric44]

Homework Statement

the arrangement in the figure

Relevant Equations

sum(F)=ma

in the opposite problem that has three pulleys and two masses, the book is asking for :
1- the tension on the string.
2- the acceleration of pulley p1.
3- the acceleration of mass m2.
4- what should the mass "m" be such that m1 does not accelerate?
note : pulley p1 has mass m, in the diagram of P1 there should be mg also along with T on pulley 1 , I assume!

the solution
1-
$$
T-m_{1}g=m_{1}a_{1},\quad,\quad T=m_{1}(g+a_{1})
$$
2-
$$
m_{2}g-T_{2}=m_{2}a_{2},\quad,\quad a_{2}=(m_{2}g-T_{2})/m_{2}
$$
3-
$$
2T-T-mg=ma_{p1}\quad,\quad a_{p1}=T/m-g
$$
4- now for a_{1} to equal zero the system will be in equilibrium and
$$
T=m1g\quad, T/m=g\quad, m=m_{1}
$$
I just want to check my answer or if I am missing something, thanks in advance.

 

[Orodruin]

Homework Statement:: the arrangement in the figure
Relevant Equations:: sum(F)=ma

note : pulley p1 has mass m, in the diagram of P1 there should be mg also along with T on pulley 1 , I assume!

This is not sufficient information about the pulley unless additional assumptions are made about the moment of inertia of that pulley.

 

[Lnewqban]

What “the opposite problem” is?

Are you sure about the pulleys having mass?

 

[patric44]

the problem shown in the figure is exactly similar problem that I found online for the free body diagram only, but my problem states the following :

 

[Orodruin]

the problem shown in the figure is exactly similar problem that I found online for the free body diagram only, but my problem states the following :
View attachment 317380

The text of this problem specifies the mass m to be on the axle of the pulley. Effectively givibg it zero moment of inertia. This makes it solvable.

 

[patric44]

The text of this problem specifies the mass m to be on the axle of the pulley. Effectively givibg it zero moment of inertia. This makes it solvable.

my bad, I miss wrote that the pulley itself has mass m, so is my approach for the solution correct

 

[Orodruin]

my bad, I miss wrote that the pulley itself has mass m, so is my approach for the solution correct

You may want to rethink this:

3-
$$
2T-T-mg=ma_{p1}\quad,\quad a_{p1}=T/m-g
$$

Which direction are you taking as positive? In which direction do the forces act?

 

[patric44]

should this be, assuming the up is positive
$$
T-2T-mg=ma_{p}\quad,\quad a_{p}=-T/m-g
$$

 

[Steve4Physics]

This overlaps previous posts of the last few minutes, but I've already written it and it might help.

Some thoughts…

You have forgotten to include the axle’s weight on the FBD for p1.

When giving each object’s ‘F=ma’ equation, you appear to be using a sign-convention where the positive direction is taken to be the assumed direction of acceleration. I think (not sure) that this will be OK - but it would mean your ‘F=ma’ equation for p1 is wrong.

You haven’t given the ‘F=ma’ equation for p2.

Since the string length is constant, there is a relationship between $a_1, a_2, a_{p1}$ and $a_{p2}$.

Edit - typo' fixed.

 

[patric44]

This overlaps previous posts of the last few minutes, but I've already written it and it might help.

Some thoughts…

You have forgotten to include the axle’s weight on the FBD for p1.

When giving each object’s ‘F=ma’ equation, you appear to be using a sign-convention where the positive direction is taken to be the assumed direction of acceleration. I think (not sure) that this will be OK - but it would mean your ‘F=ma’ equation for p1 is wrong.

You haven’t given the ‘F=ma’ equation for p2.

Since the string length is constant, there is a relationship between $a_1, a_2, a_{p1}$ and $a_{p2}$.

Edit - typo' fixed.

yes that's the convention I used, I guess it should be (I am some how confused about the direction of the acceleration) I will assume that both of them will go down ,then for P1
$$
2T+mg-T=ma_{p}\quad, a_{p}=T/m+g
$$
and for the second pulley P2
$$
T_{2}-2T=m_{p2}a_{p2}\quad, T_{2}=2T
$$

 

[haruspex]

yes that's the convention I used, I guess it should be (I am some how confused about the direction of the acceleration) I will assume that both of them will go down ,then for P1
$$
2T+mg-T=ma_{p}\quad, a_{p}=T/m+g
$$
and for the second pulley P2
$$
T_{2}-2T=m_{p2}a_{p2}\quad, T_{2}=2T
$$

Looks right so far. What about the kinematic relationship between the accelerations that @Steve4Physics mentioned?

 

[Orodruin]

$$
2T+mg-T=ma_{p}\quad, a_{p}=T/m+g
$$

This should already tell you something about the possibility of the system being in equilibrium. So this

now for a_{1} to equal zero the system will be in equilibrium and

cannot be the case. However, you may be able to find a setup such that $a_1 = 0$, but that will not mean the system is in equilibrium.

 

[patric44]

This should already tell you something about the possibility of the system being in equilibrium. So this

cannot be the case. However, you may be able to find a setup such that $a_1 = 0$, but that will not mean the system is in equilibrium.

I assumed that since m1 is not accelerating say downward, it will not lift the other mass m2 at the other side, how could m1 not be accelerating but m2 is ?

 

[patric44]

Looks right so far. What about the kinematic relationship between the accelerations that @Steve4Physics mentioned?

can you give me a little hint on how to use the length in that situation?

 

[Orodruin]

can you give me a little hint on how to use the length in that situation?

Are you familiar with how the string being a fixed length is used in the treatment of the regular Atwood machine?

 

[haruspex]

can you give me a little hint on how to use the length in that situation?

Assign variables to each straight segment of string.
Write equations representing that each whole string is of constant length.
Differentiate as necessary to get accelerations.