Choosing proper coordinates in a complex 2 pulley system

[baseballfan_ny]

Homework Statement

Block 1 and block 2, with masses m1 and m2, are connected by a system of massless,
inextensible ropes and massless pulleys as shown above.
Solve for the acceleration of block 2 in terms of m1, m2 and g. Assume that ”down” is
positive. Express your answer in terms of some or all of the following: g, m1, and m2.

Relevant Equations

F = ma
2nd derivative of length of inextensible string = 0

FBD Block 1

FBD Block 2

FBD Pulley B

I'm mainly concerned with the coordinate system direction in this problem, but just to show my attempt, here are the equations I got from the system.

$-T_A + m_1g = m_1a_1$
$T_B - m_2g = m_2a_2$
$T_A - 2T_B = 0$

Using the fact that the lengths of rope A (wrapped around pulley A) and rope B (wrapped around pulley B), are constant...

$l_A = y_1(t) + y_B(t) + Constants$
$l_A'' = a_1 + a_B =0$
$a_1 = -a_B$

$l_B = y_2(t) -2y_B(t) + Constants$
$l_B'' = a_2 -2a_B =0$
$a_B = \frac {a_2} {2}$

With 5 equations and 5 unknowns, I solved for $a_2$ and got $$a_2 = \frac {-g(2m_1 + 4m_2)} {4m_2 + m_1}$$

My main concern is with the coordinate direction in Blocks 1 and 2. The problem says to take "down" as positive, which I did for Block 1. However, for Block 2, I assumed that the positive direction would be upwards, because if block 1 were to move a bit down, Block 2 (and Pulley B) would move up. Is this the correct way of thinking about it? In an online solution of the problem, the downwards direction was chosen as positive in both Blocks 1 and 2. Appreciate your help!

Thanks in advance! (Also thought I should mention I'm new to PhysicsForums).

 

[TSny]

My main concern is with the coordinate direction in Blocks 1 and 2. The problem says to take "down" as positive, which I did for Block 1. However, for Block 2, I assumed that the positive direction would be upwards, because if block 1 were to move a bit down, Block 2 (and Pulley B) would move up. Is this the correct way of thinking about it? In an online solution of the problem, the downwards direction was chosen as positive in both Blocks 1 and 2.

I'm mainly concerned with the coordinate system direction in this problem, but just to show my attempt, here are the equations I got from the system.

You can choose the positive directions as you have chosen them. (Personally, I find the signs to be easier to handle in this problem if I follow the suggestion of the question and take positive direction to be the same for both blocks.)

$-T_A + m_1g = m_1a_1$
$T_B - m_2g = m_2a_2$
$T_A - 2T_B = 0$

These look good for your sign conventions.

Using the fact that the lengths of rope A (wrapped around pulley A) and rope B (wrapped around pulley B), are constant...

$l_A = y_1(t) + y_B(t) + Constants$
$l_A'' = a_1 + a_B =0$
$a_1 = -a_B$

$l_B = y_2(t) -2y_B(t) + Constants$
$l_B'' = a_2 -2a_B =0$
$a_B = \frac {a_2} {2}$

I believe there is an error in these equations. Are the positions $y_1(t), y_2(t)$ and $y_B(t)$ all measured from the same origin? Is $y_B(t)$ in the equation for $l_A$ the same quantity as $y_B(t)$ in the equation for $l_B$?

Suppose block 1 accelerates downwards. So, for your sign conventions, $a_1 > 0$. Your equation $a_1 = -a_B$ then implies that $a_B <0$. Then, your equation $a_B = \frac {a_2}{2}$ would imply that $a_2 <0$. For your sign convention for block 2, this would imply that block 2 accelerates downwards. So, your equations imply that if block 1 accelerates downwards, then block 2 also accelerates downwards.

With 5 equations and 5 unknowns, I solved for $a_2$ and got $$a_2 = \frac {-g(2m_1 + 4m_2)} {4m_2 + m_1}$$

One way to partially check your answer is to consider the case where neither block accelerates. It is easy to see what the relation must be between $m_1$ and $m_2$ for static equilibrium. You can then check to see if your answer for $a_2$ is zero in this case.

 

[baseballfan_ny]

Thanks for the response!

I believe there is an error in these equations. Are the positions and all measured from the same origin? Is in the equation for the same quantity as in the equation for ?

Suppose block 1 accelerates downwards. So, for your sign conventions, . Your equation then implies that . Then, your equation would imply that . For your sign convention for block 2, this would imply that block 2 accelerates downwards. So, your equations imply that if block 1 accelerates downwards, then block 2 also accelerates downwards.

I think you're right. I chose positions $y_1(t), y_B(t), and y_2(t)$ all from the top of pulley A, so I went back and made it consistent with the sign convention which got me the following:

$l_A = y_1(t) - y_B(t) + constants$
$l_A'' = a_1 - a_B = 0$
$a_1 = a_B$

$l_B = (distance from origin to floor - -y_B(t)) + constants + (-y_2(t) - -y_B(t))$
$l_B = 2y_B(t) - y_2(t) = 0$
$l_B'' = 2a_B - a_2 = 0$
$a_2 = 2a_B$

This seems to hold with the scenario of $a_1 > 0$, because $a_B > 0$ and $a_2 > 0$, which, by the sign conventions I chose would mean $a_1$ accelerates down and $a_B$ and $a_2$ accelerate upwards.

This gave me $a_2 = \frac {g(2m_1 - 4m_2)} {4m_2 + m_1}$, which held up with the test you suggested!

You can choose the positive directions as you have chosen them. (Personally, I find the signs to be easier to handle in this problem if I follow the suggestion of the question and take positive direction to be the same for both blocks.)

I did the problem again with the sign convention as the same for both blocks, and I found it much simpler. I got the same answer, except that $a_2$ had the opposite sign from the corrected answer with my convention, but I think this makes sense, because the sign convention for $a_2$ is reversed, so the sign of $a_2$ should be reversed.

I'm just a bit confused as to why you can choose the positive direction to be the same or different for both blocks. Is it because they have two separate accelerations? If the problem had one uniform acceleration for the entire system, would you have to choose a sign convention that shows the positive direction moving in "one path?" (ie the positive direction would be down on the left side of the pulley and up on the right side of the pulley).

 

[TSny]

OK. I'm still not sure I understand how you are defining $y_1(t), y_2(t)$ and $y_B(t)$. It would be helpful to indicate these in a diagram and also to specify the positive direction for each of these. But, anyway, your answer for $a_2$ looks correct.

I did the problem again with the sign convention as the same for both blocks, and I found it much simpler. I got the same answer, except that a2 had the opposite sign from the corrected answer with my convention, but I think this makes sense, because the sign convention for a2 is reversed, so the sign of a2 should be reversed.

Sounds good.

I'm just a bit confused as to why you can choose the positive direction to be the same or different for both blocks. Is it because they have two separate accelerations? If the problem had one uniform acceleration for the entire system, would you have to choose a sign convention that shows the positive direction moving in "one path?" (ie the positive direction would be down on the left side of the pulley and up on the right side of the pulley).

No, you would not have to choose the positive direction to be the same for both blocks.

Suppose block 1 rests on top of block 2 and you accelerate both blocks vertically upward with a force F.

If you draw free body diagrams for each block, you do not need to take the positive direction of acceleration to be the same for the blocks. (However, you would be unnecessarily complicating things in this case if you chose different positive directions!) If you choose upward as positive for block 1 and downward as positive for block 2, then as long as you consistently apply Newton's laws, you will be able to determine the correct acceleration of each block and the correct value for the normal force between the blocks. This would be an odd way to do it, and I would not recommend it (unless you want to drive your professor crazy ).

 

[Lnewqban]

... I'm just a bit confused as to why you can choose the positive direction to be the same or different for both blocks. Is it because they have two separate accelerations? If the problem had one uniform acceleration for the entire system, would you have to choose a sign convention that shows the positive direction moving in "one path?" (ie the positive direction would be down on the left side of the pulley and up on the right side of the pulley).

In cases like this, I eliminate the change of direction that the pulleys introduce.
I turn the schematic sideways and unwrap the ropes from around the pulleys.

By doing that, I have:
A block 1 with force $m_1g$ pulling toward the left.
A straight rope connecting block 1 and middle point of lever B (which has a fulcrum anchored to the floor and another straight rope attached to its free end).
Another straight rope connecting free end of lever B and block 2.
A block 2 with force $m_2g$ pulling toward the right.

The whole system (which floats horizontally in my imagination) has a net resultant force acting upon it and corresponding accelerations, all pointing in the same direction.

 

[baseballfan_ny]

OK. I'm still not sure I understand how you are defining and . It would be helpful to indicate these in a diagram and also to specify the positive direction for each of these. But, anyway, your answer for looks correct.

Ahh, sorry about the confusion. I attached a diagram below of how I had defined $y_1(t), y_B(t),$ and $y_2(t)$ using the sign convention I originally did for reference of any viewers. Glad to here that the answer checks out though!

If you draw free body diagrams for each block, you do not need to take the positive direction of acceleration to be the same for the blocks. (However, you would be unnecessarily complicating things in this case if you chose different positive directions!) If you choose upward as positive for block 1 and downward as positive for block 2, then as long as you consistently apply Newton's laws, you will be able to determine the correct acceleration of each block and the correct value for the normal force between the blocks. This would be an odd way to do it, and I would not recommend it (unless you want to drive your professor crazy ).

That example cleared it up, thanks! I tried the example using both a normal sign convention and an "opposite" one, and though I got consistent answers, I 100% agree that it's much easier to choose a consistent sign convention!

Thank you so much for your help!

 

[baseballfan_ny]

In cases like this, I eliminate the change of direction that the pulleys introduce.
I turn the schematic sideways and unwrap the ropes from around the pulleys.

By doing that, I have:
A block 1 with force $m_1g$ pulling toward the left.
A straight rope connecting block 1 and middle point of lever B (which has a fulcrum anchored to the floor and another straight rope attached to its free end).
Another straight rope connecting free end of lever B and block 2.
A block 2 with force $m_2g$ pulling toward the right.

The whole system (which floats horizontally in my imagination) has a net resultant force acting upon it and corresponding accelerations, all pointing in the same direction.

That's a cool way to think about it! I tried it like that and got the same equations from the FBDs.