2.2-8 f A rectangular block has mass of 100 kg is resting on an inclined sur->J

Question

2.2-8 f A rectangular block has mass of 100 kg is resting on an inclined sur->J face, as in the figure. If the static coefficient of friction between the block and the surface is ps = 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 = 0.2, find the magnitude of the force F needed to sustain the upward sliding motion of the block under constant velocity. The bar shown in the figure below is supported by a roller at point

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* sin(30) = m *g * sin(theta) - us * (m * g * cos(theta) + F * sin(theta))

F = 100 * 9.8 *sin(30 degree) - 0.25 * ( 100 * 9.8 * cos(30 degree) + F * sin(30 degree))

F = 246.9 N

the force needed to prevent sliding is 246.9 N

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for starting the upwards motion

F = m *g * sin(theta) + us * m * g * cos(theta) + us *F * sin(theta)

F = 100 * 9.8 *(sin(30 degree) + 0.25 * cos(30 degree)) + 0.25 * F * sin(30 degree)

F = 802.2 N

the force needed to move upwards is 802.2 N

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for the force needed to sustain the motion

F = m *g * sin(theta) + uk * m * g * cos(theta) + uk *F * sin(theta)

F = 100 * 9.8 *(sin(30) + 0.2 * cos(30)) + 0.2 * F * sin(30 degree)

F = 733.05 N

the force needed to move upwards is 733.05 N

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