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.
Torque= (radius)(force)sin(angle between them)= (inerta)*(alpha)or (inertia)*a/r
friction= (mew)*(normal force)
The Attempt at a Solution
mo= mass of pulley
I think this problem just has too much algebra =.=
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)
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!