A block weighing 82.5 N rests on a plane inclined at 25.0° to the horizontal. A

Question

A block weighing 82.5 N rests on a plane inclined at 25.0° to the horizontal. A force F is applied to the object at 40.0° to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.374 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

m*g = 82.5 N

(a)
F*cos(15) + Fs = m*g*sin(25)
F*cos(15) + 0.374*[82.5*cos(25) - F*sin(15)] =  82.5*sin(25)
F = 7.94 N

Minimum Force F that will prevent the block from slipping down the plane, F = 7.94 N

(b)
F*cos(15) - Fs = m*g*sin(25)
F*cos(15) - 0.374*[82.5*cos(25) - F*sin(15)] =  82.5*sin(25)
F = 59.12 N
Minimum Force F that will start the block moving up the plane, F = 59.12 N

(c)
F*cos(15) = m*g*sin(25) + Fk
F*cos(15) = 82.5*sin(25) + 0.156 *[82.5 * cos(25) - F*sin(15)]
F = 46.2 N
Value of F will move the block up the plane with constant velocity, F = 46.2 N

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