The first mass M (9.50 kg) is sitting on an inclined plane whose angle above the

Question

The first mass M (9.50 kg) is sitting on an inclined plane whose angle above the horizontal is 35 degrees. This mass is connected by a massless cord over a massless and frictionless pulley to the second mass m. The coefficient of static friction is 0.530 and the coefficient of kinetic friction is 0.472 for the block on the incline.

(a) What is the minimum mass, m, in kg, needed to keep the block from sliding down the incline?

(b) What mass m, in kg, would be needed to make the block move up the incline at a constant speed?

Explanation / Answer

Part A)

For the hanging mass we have mg = T

For the mass on the incline we have T + uMGcos 35 = Mgsin(35)

Combine the two and get mg = Mgsin35 - uMgcos(35)

m(9.8) = (9.5)(9.8)(sin 35) - (.53)(9.5)(9.8)(cos 35)

m = 1.32 kg

Part B)

Same basic idea, but.

For the mass on the incline we have T = uMGcos 35 + Mgsin(35)

Combine the two and get mg = Mgsin35 + uMgcos(35)

m(9.8) = (9.5)(9.8)(sin 35) + (.472)(9.5)(9.8)(cos 35)

m = 9.12 kg

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