A block P of mass 3 13 kg is lying on a rough inclined plane of angle theta = 0.

Question

A block P of mass 3 13 kg is lying on a rough inclined plane of angle theta = 0.75 radians Block P is attached via a model suing 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 = 0.36 Calculate the value of the mass of block Q. in kg. that is just sufficient to initiate the sliding motion of block Q up the inclined plane Give your answer to 3 decimal places m

Explanation / Answer

Let the mass of Q =M

Mass of P = m = 3.13 Kg

Therefore

Mg - µmg - mgsin = 0

Therefore

M = µm + msin

M = m(µ + sin)

M = 3.13*(0.36 + sin((360/2 * 0.75))

M = 3.26 Kg

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