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## Homework Statement

Consider the setup shown below. The blocks

have masses 8.2 kg and 25 kg. The pulley

has mass 5.9 kg, and is a uniform disc with

radius 0.25 m. Assume the pulley to be fric-

tionless, but the coefficient of friction between

the block and the surface is 0.28.What is the

acceleration of the blocks? Assume the 25 kg

mass is descending with acceleration a. The

moment of inertia of the disk is 1/2MR2 and the

acceleration of gravity is 9.8 m/s2 .

Answer in units of m/s2.

## Homework Equations

Torque= (radius)(force)sin(angle between them)= (inerta)*(alpha)or (inertia)*a/r

sumF=ma

friction= (mew)*(normal force)

## The Attempt at a Solution

mo= mass of pulley

I think this problem just has too much algebra =.=

sumF1x= m1gsin(theta)+Friction-T1=m1a

SumF2y=T2-m2g=m2a

sumT= 1/2(mo)ra= rt1-rt2 ( I don't know which one should go first or why)

solving the t and plugging it into torque I receive:

a= g( (1+(mew))m1sin(theta)-m2)

-----------------------------

(1/2mo+m1+m2)

I have a practice problem I can use to verify if i set this problem up but I keep getting like .05 off -.-

Any help would be appreciated!

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