Double Pulley Problem: Relations Between Masses and Accelerations

[jessedevin]

Homework Statement

An object of mass m1 on a frictionless horizontal table is connected to an object of mass m2 through a very light pulley P1 and a light fixed pulley P2.

(a) If a1 and a2 are the accelerations of m1 and m2, respectively, what is the relation between these accelerations?

(b) Express the tensions in the strings in terms of g and the masses m1 and m2.

(c) Express the accelerations a1 and a2 in terms of g and the masses m1 and m2.

Homework Equations

\SigmaF = ma

The Attempt at a Solution

The way that I did it was I made two force diagrams. The first force diagram of m1 I made just has force going to the right, force T1 (tension of rope). Then I have a second force diagram of m2 with a force T2 going up and a force of mg (or weight of m2) going down. but the thing is that I keep on getting one a (acceleration), but the answer has a1 and a2. Can someone please tell me what I am doing wrong with my force diagrams. I think if i can have the correct force diagram, then I will be able to do the problem (hopefully!).

 

[LowlyPion]

Homework Statement

An object of mass m1 on a frictionless horizontal table is connected to an object of mass m2 through a very light pulley P1 and a light fixed pulley P2.

(a) If a1 and a2 are the accelerations of m1 and m2, respectively, what is the relation between these accelerations?

(b) Express the tensions in the strings in terms of g and the masses m1 and m2.

(c) Express the accelerations a1 and a2 in terms of g and the masses m1 and m2.

Homework Equations

\SigmaF = ma

The Attempt at a Solution

The way that I did it was I made two force diagrams. The first force diagram of m1 I made just has force going to the right, force T1 (tension of rope). Then I have a second force diagram of m2 with a force T2 going up and a force of mg (or weight of m2) going down. but the thing is that I keep on getting one a (acceleration), but the answer has a1 and a2. Can someone please tell me what I am doing wrong with my force diagrams. I think if i can have the correct force diagram, then I will be able to do the problem (hopefully!).

The movable pulley gives you a mechanical advantage of 2, but it means the tension translated to m1 is only 1/2 from T2, and hence so too will the acceleration of m1 be different.

Stated as Tensions that's T2 = 2T1

 

[jessedevin]

The movable pulley gives you a mechanical advantage of 2, but it means the tension translated to m1 is only 1/2 from T2, and hence so too will the acceleration of m1 be different.

Stated as Tensions that's T2 = 2T1

Okay, so if T2= 2T1, then the force diagram looks like this (this is what i got):
The force diagram of m1 has T1 going to the right, m1g going down, and the normal vector n going up.
The force diagram of m2 has T2 going up and m2g going down.

So here's the way I did it, but its wrong for some reason, and i don't no why:
\SigmaFm1= T1 = m1a1
\SigmaFm2= T2 - m2g=m2a2.

So what i did to find the relationship between the accelerations was substitute T2 = 2T1.
2m1a1 - m2g= m2a2
a1= m2(a2 + g)/2m1
But it says that i am wrong, and there is not suppose to be the gravity constant in the relationship between the accelerations. Can someone please tell me what am i doing wrong?

 

[LowlyPion]

But it says that i am wrong, and there is not suppose to be the gravity constant in the relationship between the accelerations. Can someone please tell me what am i doing wrong?

(b) Express the tensions in the strings in terms of g and the masses m1 and m2.

(c) Express the accelerations a1 and a2 in terms of g and the masses m1 and m2.

You can't avoid g.

Edit: But in part (a) the relationship of the accelerations falls out of the pulley doesn't it?

 

[jessedevin]

You can't avoid g.

I mean for part (a)

 

[jessedevin]

Should i determine part (b) or (c) before I do (a)

 

[LowlyPion]

I mean for part (a)

Sorry, I edited my previous post when I realized you were talking about a).

Here it is again. But in part (a) the relationship of the accelerations falls out of the pulley doesn't it?

 

[jessedevin]

Sorry, I edited my previous post when I realized you were talking about a).

Here it is again. But in part (a) the relationship of the accelerations falls out of the pulley doesn't it?

What do you mean? That a1=a2? I am really confused...

 

[LowlyPion]

What do you mean? That a1=a2? I am really confused...

Visualize how far you have to pull the rope at the m1 side of the pulley to move the weight on the m2 side. Don't you have to move a distance of 2 for every 1 you move on the m2 side? Isn't that where the mechanical advantage comes from? You pull only 1/2 as much, but over twice the distance to move things on the m2 side? For any velocity at m2 then there is 2 times the velocity at m1.

 

[jessedevin]

Okay i get that for any velocity at m2 there is 2 times the velocity at m1, so a1= 2a2. But I am confused on how to set of the force equations.Did i write the force equations right? Can you give me a hint on how to set it up?

Force eqns:
\Sigma Fm1= T1 = m1a1
\Sigma Fm2= T2 - m2g=m2a2.

 

[LowlyPion]

\SigmaFm1= T1 = m1a1
\SigmaFm2= T2 - m2g=m2a2.

So what i did to find the relationship between the accelerations was substitute T2 = 2T1.

Ok, one equation is wrong. T2 is minus and M2*g is plus as you need to preserve the same sign convention between the two equations if you want to solve them.

 

[LowlyPion]

Okay i get that for any velocity at m2 there is 2 times the velocity at m1, so a1= 2a2. But I am confused on how to set of the force equations.Did i write the force equations right? Can you give me a hint on how to set it up?

Force eqns:
\Sigma Fm1= T1 = m1a1
\Sigma Fm2= T2 - m2g=m2a2.

If 2x1 = x2
and 2v1 = v2
What is a1 again?

 

[jessedevin]

If 2x1 = x2
and 2v1 = v2
What is a1 again?

a1=2a2 ...lol

 

[jessedevin]

Ok, one equation is wrong. T2 is minus and M2*g is plus as you need to preserve the same sign convention between the two equations if you want to solve them.

how is that eqn wrong. m2g is the force going down (hence, negative) and T2 is going up, isn't the force eqn i wrote right? Also can you give me any more hints on writing each tension T in terms of m1, m2 and gravity?

 

[LowlyPion]

how is that eqn wrong. m2g is the force going down (hence, negative) and T2 is going up, isn't the force eqn i wrote right? Also can you give me any more hints on writing each tension T in terms of m1, m2 and gravity?

You need to make Tension the same sign in whatever direction you choose.

You chose to the right in equation 1 and it translates to the left in equation 2. As the problem stands it is net down. So T must be < m2g

 

[jessedevin]

Okay i am still drawing a blank on this problem... So let's go back from the beginning. My textbook does not explain mechanical advantage, so could you please explain a little more in depth on mechanical advantage on having two pulleys in relation to the two masses?

 

[LowlyPion]

Okay i am still drawing a blank on this problem... So let's go back from the beginning. My textbook does not explain mechanical advantage, so could you please explain a little more in depth on mechanical advantage on having two pulleys in relation to the two masses?

Maybe this will help?
http://en.wikipedia.org/wiki/Pulley#Types_of_systems

 

[jessedevin]

It helped a little, but still kind of foggy. Is there any more hints you could give me to help me solve the problem? I am completely stuck.

 

[jessedevin]

Can anyone give me more hints to answer this problem?

 

[astronomerc]

Sorry, I edited my previous post when I realized you were talking about a).

Here it is again. But in part (a) the relationship of the accelerations falls out of the pulley doesn't it?

I think, we set the T's equal , solve for a1 which will give us a2. I don't think the a's fall out when solving for the tension, but I think we can find the a's... I hope.