A block P of mass 3.86 kg is lying on a rough inclined plane of angle theta = 0.

Question

A block P of mass 3.86 kg is lying on a rough inclined plane of angle theta = 0.87 radians. Block P is attached via a model string that passes over a model pulley to a second block Q of unknown mass M kg hanging vertically as shown in diagram. The coefficient of static friction between the block P and the inclined plane is mu = 0.27. Calculate the value of the mass of block Q, in kg, that is just sufficient to initiate the sliding motion of block P up the inclined plane. Give your answer to 3 decimal places.

Explanation / Answer

theta = 0.87 radians = 49.85°

Using F = ma on P

T - 3.86 * g sin 49.85 - Fmax = 0

Using F = ma on Q

M g - T = 0 , T = M g

M g - [3.86 * g sin 49.85] - Fmax = 0

Fmax = mu * R = 0.27 (3.86 * g cos 49.85)

Fmax = 1.04 * g * cos 49.85

M g - [3.86 * g sin 49.85] - [1.04 * g * cos 49.85 ] = 0

M = 3.62 kg

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