A block weighing 77.5 N rests on a plane inclined at 25.0° to the horizontal. A

Question

A block weighing 77.5 N rests on a plane inclined at 25.0° to the horizontal. A force F is applied to the object at 35.0° to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.389 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

force due to gravity = mgsin25 = 77.5sin25 = 32.75N

maximum force due to friction = smgcos = 0.389*77.5*cos25 = 27.32N

when F is applied, maximum frictional force = s(mgcos-Fsin10) = 27.32-0.0675F

a) 27.32 - 0.0675F + Fcos10 = 27.32 - 0.0675F + 0.9848F = mgsin25

F = (32.75-27.32)/0.9173 = 5.92N

b) 27.32 - 0.0675F + mgsin25 = Fcos10 = 0.9848F

F = (27.32+32.75)/1.0523 = 57.08N

c) k(mgcos - Fsin10) + mgsin25 = Fcos10 = 0.9848F

10.96 - 0.027F + 32.75 = 0.9848F

F = 43.71/1.01 = 43.27N

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