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## Homework Statement

A string passing over a pulley has a 3.80 kg mass hanging from one end and a 3.15 kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 0.04 m and mass 0.80 kg.

It is found that if the heavier mass is give a downward speed of 0.20 m/s, it comes to rest in 6.2 s. What is the average frictional torque acting on the pulley?

## Homework Equations

T = F*r

F = ma

T = I* alpha

## The Attempt at a Solution

Newton's second law for masses:

F

_{TA}- m

_{a}g= m

_{a}*a (the heavier one)

m

_{b}*g -F

_{TB}= m

_{b}*a

For the pulley:

F

_{TB}*r

_{pulley}(ccw-positive) - F

_{TA}*r

_{pulley}(cw-negative) + (+-_____ )= I*alpha

The pulley is rotating ccw, so I thought I had to write minus friction because the friction must be opposite of rotation direction, but then saw this and got really confused.

https://www.physicsforums.com/threa...que-acting-on-the-pulley.232961/#post-1716480

Determining the sign of frictional torque in this equation, should it be positive or negative?