A block weighing 70.0 N rests on a plane inclined at 25.0° to the horizontal. A

Question

A block weighing 70.0 N rests on a plane inclined at 25.0° to the horizontal. A force F is applied to the object at 45.0° to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.388 and 0.156.

(a) What is the minimum value of F that will prevent the block from slipping down the plane?

(b) What is the minimum value of F that will start the block moving up the plane?

(c) What value of F will move the block up the plane with constant velocity?

Explanation / Answer

Here ,

a)

here ,

Normal force , N = mg * cos(25) - F * sin(20)

for preventing the block from slipping

mg * sin(25)- F * cos(20) - u * N - 0

70 * sin(25) - F * cos(20) - 0.388 * (70* cos(25) - F * sin(20)) = 0

solving for F

F = 6.2 N

the force F needed is 6.2 N

b)

For block to move up the plane

F * cos(20)- u N - mg * sin(25) = 0

-70 * sin(25) + F * cos(20) - 0.388 * (70* cos(25) - F * sin(20)) = 0

solving for F

F = 50.4 N

the force F needed is 50.4 N

c)

for moving the block with constant speed up the ramp

F * cos(20)- kinetic friction coefficient * N - mg * sin(25) = 0

-70 * sin(25) + F * cos(20) - 0.156 * (70* cos(25) - F * sin(20)) = 0

solving for F

F = 39.7 N

the force F needed is 39.7 N

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