A block weighing 70.0 N rests on a plane inclined at 25.0 degree to the horizont

Question

A block weighing 70.0 N rests on a plane inclined at 25.0 degree to the horizontal. A force F is applied to the object at 45.0 degree to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.389 and 0.156. What is the minimum value of F that will prevent the block from slipping down the plane? What is the minimum value of F that will start the block moving up the plane? What value of F will move the block up the plane with constant velocity?

Explanation / Answer

For prevention of slipping,

45 degrees to horizontal on an inclination of 25 degrees.

So force is applied at 70 degrees to the inclined plane

F cos 70 - mg sin(25) + mu * N = 0

N = mg*cos(25) + F sin(70)

F cos70 - mgsin(25) + mu*(mg cos(25) + F sin(70)) = 0

F = (70*sin(25) - 0.389* 70*cos(25)) /(cos(70) +0.398*sin(20)) = 10.25 N

For block to move up

F cos 70 - mg sin(25) - mu * N = 0

N = mg*cos(25) + F sin(70)

F cos70 - mgsin(25) - mu*(mg cos(25) + F sin(70)) = 0

F = (70*sin(25) +0.389* 70*cos(25)) /(cos(70) -0.398*sin(20)) = 263.54 N

For a constant velocity

F = (70*sin(25) +0.156* 70*cos(25)) /(cos(70) -0.156*sin(20)) = 136.78 N

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