A loaded penguin sled weighing 86.0 N rests on a plane inclined at angle theta =

Question

A loaded penguin sled weighing 86.0 N rests on a plane inclined at angle theta = 21.0 degree to the horizontal (see the figure). Between the sled and the plane, the coefficient of static friction is 0.280, and the coefficient of kinetic friction is 0.140. (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?

Explanation / Answer

(a) If you drew the free body diagram correctly, you would see that there are three forces acting on the sled.
By Newton's Second Law,
F(net) = ma = F(applied) + F(friction) - F(gravity)
Weight = mg, F(friction) = mgucos(theta) and F(gravity) = mgsin(theta)
Take two notes:
1. If the sled is to not slip as a result of all three forces coming together, it will not move at all and its acceleration will be zero.
2. If the sled is said to not move, it is subject to static friction. Therefore,
0 = F(applied) + 86(0.28)cos(21) - 86*sin(21)
F(applied) = 8.34 N

(b) The applied force in this case must overcome both the force of gravity and the force of static friction; therefore,
F(applied) = F(gravity) + F(friction)
F(applied) = 86*sin(21) + 86(0.28)cos(21) = 53.3 N

(c) If the sled is to move at a constant velocity, then its acceleration must be zero.
Also, the applied force is opposing the force of gravity and the force of kinetic friction given that the force has overcome the force of static friction, therefore:
F(applied) = F(friction) + F(gravity)
F(applied) = 86(0.14)cos(21) + 86sin(21) = 42.06 N

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