Consider the setup shown below. The blocks have masses 8.4 kg and 25 kg. The pulley has mass 6.1 kg, and is a uniform disc with ?

Question

Consider the setup shown below. The blocks have masses 8.4 kg ( on the ramp) and 25 kg.(vertically hanged)  The pulley has mass 6.1 kg, and is a uniform disc with radius 0.17 m. Assume the pulley to be frictionless, but the coefficient of friction between the block and the surface is 0.27.
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 2 M R2 and the acceleration of gravity is 9.8 m/s2 . Answer in units of m/s2. theta=31

Answer 1

motive force MF = g*25

opposing force OF = g*8.4*(sin 31+cos 31*μ )

equivalent mass of the pulley mpe = mp/2 = 3.05 kg 

acceleration a = 9.8*(25-8.4*(sin 31+cos 31*0.27 ))/(25+8.4+3.05) = 5.04 m/sec^2

Answer 2

Find the driving forces.(8.4*sin(31) + 25) * g  then subtract friction

8.4 *cos(31)*0.27*g

This force accelerates the masses of 8.4, 25 and half of the pulley's mass 3.05  -> acceleration= (8.4*sin(31)- 8.4*cos(31)*0.27 +25)*9.8 / 36.45



You can understand that as the moment of inertia is 1/2 m r^2 that it is effectively rotating half the mass.  ie it takes the same energy as moving half the mass to the same peripheral speed.

Answer 3

Provide the diagram, as the angle of ramp is also required.