# Pulleys and multiple weights

brake4country

## Homework Statement

A 100 kg man dangles a 50 kg mass from the end of a rope. If he stands on a frictionless surface and hangs the mass over a cliff with a pulley, the tension in the rope will be:
(A) 250 N
(B) 333 N
(C) 500 N
(D) 667 N

The answer is B but I am having a difficult time setting this one up.

F=ma

## The Attempt at a Solution

Okay, so we know that for the man, T = ma because he is on a frictionless surface and the block is mg = T+ma.
Some others have posted this problem on this site but I need clarification for the second equation listed above. I cannot combine these equations because the masses are different. So, what I did was alter the equations as

man: T=2ma
block: mg = T+ma

substituting gives me: mg = 2ma +ma
therefore, g = 3a and a = 3.33 m/s

But when I go back to my "altered" equations to solve for T, I get 667 N which is wrong. Anyone know what I am doing wrong here?

Brian T
Well, if the acceleration on the man is 3.33 m/s, and the only force acting on him is the tension, then you that the force of tension is his mass times that acceleration. I think the issue you had was that you plugged it back into your equation T=2ma but you used the mass for the man. At that point 2m for m block = m of the man. So when you plug it back, make sure you use the block's mass. T = 2(50)(3.33) which is B.

When you are dealing with multiple masses (or any of the same variables for that matter), be clear to label them so that you do not mix them up.

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brake4country
Ok, conceptually that makes sense. I wouldn't be using the man's mass because at that point when he is sliding, only the force due to gravity is acting on the block.

brake4country
Ok, so using the man's mass would be erroneous. To find the tension, I would have to use the mass of the block: T = 2ma = 2(50)(3.33) = 333N. The way I am thinking about this problem is that the tension is created by the object that is being affected by gravity.