A loaded penguin sled weighing 80 N rests on a plane inclined at angle =20 degre

Question

A loaded penguin sled weighing 80 N rests on a plane inclined at angle =20 degress to the horizontal. between the sled and the plane, the coefficent of kinetic of static friction is .25. and the coefficent of kinetic friction is .15 (a) what is the least magnitude of the force F, parallal to the plane what 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

Weight of the sled W = 80 N Mass of the sled m = W/g = 80 N/(9.8 m/s^2)                                            = 8.163 Kg Angle of inclination = 20 deg Coefficient of static friction s = 0.25 Coefficient of kinetic friction k = 0.15 (a) Force F = mg[ sin - s cos]              = 80[sin20 - 0.25 * cos20]              = 8.567 N (b) Force F = mg[ sin + s cos]              = 80[sin20 + 0.25 * cos20]              = 46.15 N (c) Force F = mg[ sin + k cos]              = 80[sin20 + 0.15 * cos20]              = 38.64 N

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