Chapter 06, Problem 016 A loaded penguin sled weighing 82.0 N rests on a plane i

Question

Chapter 06, Problem 016 A loaded penguin sled weighing 82.0 N rests on a plane inclined at angle 22.0° to the horizontal (see the figure). Between the sled and the plane, the coefficient of static friction is 0.300, and the coefficient of kinetic friction is 0.120. (a) What is the minimum magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane? (b) What is the minimum magnitude F that will start the sled moving up the plane? (c) What value of F is required to move the sled up the plane at constant velocity? (a) Number b) Number (c) Number Units Units Units

Explanation / Answer

a)Weight of sled = 82 N

static friction force acting on sled = 0.3*82*cos22 = 22.81 N

Let F be the minimum force required to prevent the sled from sliding

Therefore, F = 82sin22 + 22.81=30.72+22.81=53.53 N

b) The minimum force that will be required to start the sled to move is also same as the force obtained in part (a) = 53.53N

c)When the sled will be moving at constant speed, sled will be subjected to kinetic friction force

kinetic friction force = 0.12*82*cos22 = 9.12 N

Let the force required to move the sled at constant speed be F

F = 82sin22 + 9.12 =39.84 N

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