A loaded penguin sled weighing 64 N rests on a plane inclined at 20 degree to th

Question

A loaded penguin sled weighing 64 N rests on a plane inclined at 20 degree to the horizontal. Between the sled and the plane the coefficient of static friction is 0.27, and the coefficient of kinetic friction is 0.14. (a) What is the minimum magnitude of the force F vector, parallel to the plane, that will prevent the sled from slipping down the plane? N (b) What is the minimum magnitude F that will start the sled moving up the plane? N (c) What value of F is required to move the block up the plane at constant velocity?

Explanation / Answer

Part A)

From the sum of forces on the block along the plane...

Fa + Ff = Fw(sin angle)

Fa + umgcos(angle) = mgsin(angle)

Fa + .28(64)cos(20) = (64)(sin 20)

Fa = 5.05 N

Part B)

Fa = Ff + Fw(sin angle)

Fa = umgcos(angle) + mgsin(angle)

Fa = .28(64)cos(20) + (64)(sin 20)

Fa = 38.7 N

Part C)

Fa = Ff + Fw(sin angle)

Fa = umgcos(angle) + mgsin(angle)

Fa = .14(64)cos(20) + (64)(sin 20)

Fa = 30.3 N

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