A rectangular block has mass of 100 kg is resting on an inclined surface, as in
ID: 1593046 • Letter: A
Question
A rectangular block has mass of 100 kg is resting on an inclined surface, as in the figure. If the static coefficient of friction between the block and the surface is mu_s = 0.25, find the magnitude F of the horizontal force needed to prevent its downward sliding. Also, find the magnitude F of the horizontal force needed to initiate an upward sliding of the block. If the kinetic coefficient of friction is /j* = 0.2, find the magnitude of the force F needed to sustain the upward sliding motion of the block under constant velocity.Explanation / Answer
Here ,
mass , m = 100 Kg
coefficient of frictional force , us = 0.25
coefficient of kinetic friction, uk = 0.20
force needed to prevent downward sliding
F = m *g * sin(theta) - us * m * g * cos(theta)
F = 100 * 9.8 *(sin(30) - 0.25 * cos(30))
F = 277.8 N
the force needed to prevent sliding is 277.8 N
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for starting the upwards motion
F = m *g * sin(theta) + us * m * g * cos(theta)
F = 100 * 9.8 *(sin(30) + 0.25 * cos(30))
F = 702.2 N
the force needed to move upwards is 702.2 N
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for the force needed to sustain the motion
F = m *g * sin(theta) + uk * m * g * cos(theta)
F = 100 * 9.8 *(sin(30) + 0.2 * cos(30))
F = 659.7 N
the force needed to move upwards is 659.7 N