Two masses and a pulley...

Oct 2008
66
13
Hey, was wondering if anyone might want to help me check my thinking on this fairly basic exercise? We have a system consisting of a pulley in the form of a solid cylinder and two blocks hanging from it by a massless rope. The goal is to find the total force on the string by which the pulley is attached to the roof.

Variables:
The masses of the blocks are
mA and mB, where mA < mB.
mA hangs on the left hand side, mB hangs on the right hand side.

The mass and radius of the pulley are
mC and rC, respectively
and thus its moment of inertia is I = (mC * rC^2)/2

The gravitational acceleration is
g

The outline of my solution goes like this:

Three separate forces contribute to the total force:
The weight of the pulley
The forces in the rope on both sides of the pulley, which I denote
TA and TB

I'm assuming that the rope doesn't slip, and thus the magnitude of the acceleration is the same on both sides, as well as tangentially along the pulley. I denote this acceleration
aY

I've chosen my coordinate axes to be positive to the right, up and out of the paper. Thus the block pulling in the counterclockwise direction should give a positive moment and vice versa, right?

Give the above, I want to pose the solution like this:

I*(aY / r) = TA*r - TB*r (Newton's second for rotational motion)
mA*aY = TA - mA*g (Newton's second for block A)
mB*aY = mB*g - TB (Newton's second for block B)

Three unknowns, three equations. Seems like smooth sailing, but I get the wrong answer. I suspect I'm simply doing a slip-up somewhere in the calculation, but if anyone can spot some misunderstanding in how I approach the problem, or maybe even confirm that I'm on the right track, I'd be very grateful!
 

ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
You need to be very careful with signs. With your equations:

I*(aY / r) = TA*r - TB*r (Newton's second for rotational motion)
mA*aY = TA - mA*g (Newton's second for block A)
mB*aY = mB*g - TB (Newton's second for block B)

the sign of aY is inconsistent. In the first equation you get a positive value for aY if TA>TB, meaning the pulley rotates counter-clockwise (assuming mA is on the left side of the pulley), and mA is falling. But in the 2nd equation a positive value of aY comes about only if TA>mA*g, and hence mA is rising. You need a minus sign in front of the left hand side of that equation. The third equation has the same issue - positive aY would mean that TB>mB*g, but in that case the right hand side is negative, so again you need a negative sign in front of the left hand side. Try it and see if this helps. Also, despite your stated convention of positive always being up, I assume the value of 'g' you are using is a positive number, not negative, right?
 
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