A loaded penguin sled weighing 60 N rests on a plane inclined at angle theta = 2

Question

A loaded penguin sled weighing 60 N rests on a plane inclined at angle theta = 24 degree to the horizontal (see the figure). Between the sled and the plane, the coefficient of static friction is 0.26, and the coefficient of kinetic friction is 0.14. What is the minimum magnitude of the force vector F, parallel to the plane, that will prevent the sled from slipping down the plane? What is the minimum magnitude That will start the sled moving up the plane? What value of F is required to move the sled up the plane at constant velocity?

Explanation / Answer

Weight of the sled W = 60 N

Mass of the sled m = W/g = 60 N/(9.8 m/s^2) = 6.116 kg

Angle of inclination = 240

Coefficient of static friction s = 0.26

Coefficient of kinetic friction k = 0.14

(a)

Force F = mg[ sin - s cos]

             = 60[sin24 - 0.26 * cos24]

             = 10.15 N

(b)

Force F = mg[ sin + s cos]

             = 60[sin24 + 0.26 * cos24]

             = 38.65 N

(c)

Force F = mg[ sin + k cos]

             = 60[sin24 + 0.14 * cos24]

             = 32.07 N

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