A block weighing 67.5 N rests on a plane inclined at 25.0degree to the horizonta

Question

A block weighing 67.5 N rests on a plane inclined at 25.0degree to the horizontal. A force F is applied to the object at 40.0degree to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.400 and 0.156. (a) What is the minimum value of F that will prevent the block from slipping down the plane? N (b) What is the minimum value of F that will start the block moving up the plane? N (c) What value of F will move the block up the plane with constant velocity? N

Explanation / Answer

here,

weight of block , w = 67.5 N

theta = 25 degree

phi = 40 degree

coefficient of static friction , us = 0.4

coefficient of kinetic friction ,uk = 0.156

(a)

let the minimum value of force be F

normal reaction , N = w * cos(theta) - F * sin(phi)

N = 67.5 * cos(25) - F * sin(40)

N = 61.16 - F * 0.64 .....(1)

and

for the block to not slide

- F * cos(phi) + w * sin(theta) - us * N = 0

- F * cos(40) + 67.5 * sin(25) - 0.4 * ( 61.6 - F * 0.64) = 0

F = 7.63 N

(b)

for the block to move up the incline

F * cos(phi) - w * sin(theta) - us * N = 0

F * cos(40) - 67.5 * sin(25) - 0.4 * ( 61.6 - F * 0.64) = 0

F = 52.02 N

(c)

for the block to move up the incline with constant velocity

F * cos(phi) - w * sin(theta) - uk * N = 0

F * cos(40) - 67.5 * sin(25) - 0.156 * ( 61.6 - F * 0.64) = 0

F = 44.02 N

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