A block weighing 72.5 N rests on a plane inclined at 25.0° to the horizontal. A

Question

A block weighing 72.5 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, 0.333 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

Resolving Forces we get

Fup= FCos (30-20) = FCos 10 =0.98 F

Fperpendicular = F Sin (30-20) = F Sin10

Normal Reaction = 72.5 Cos 25 - FSin 10 = 65.71 - 0.17F

Gravitational force down the incline = 72.5 Sin 25 = 30.64 N

Frictional force

fstatic=us N = 0.333 *(65.71 - 0.17F)

fdynamic=ud N = 0.156*(65.71 - 0.17F)

We get

(a) For this case frictional will act upwards

Balanceing forces

0.98 F + 0.333 *(65.71 - 0.17F) = 30.64 N

Solving we get F=9.44 N

(b) In this case fstatic will act downwards

0.98 F - 0.333 *(65.71 - 0.17F) = 30.64 N

F=50.67 N

(c) In this case fdynamic will act downwards

0.98 F - 0.156 *(65.71 - 0.17F) = 30.64 N

F= 40.63 N

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