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 0.30, and the coefficient of kinetic friction is 0.17. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane? What is the minimum magnitude F that will start the sled moving up the plane? What value of F is required to move the block up the plane at constant velocity?

Explanation / Answer

given,

weight of the penguin = 64 N

angle = 20 degree

coefficient of friction = 0.3

coefficient of kinetic friction = 0.17

if we apply force F and block is in rest so force equation of force

mg * sin(20) = F + 0.3 * mg * cos(20)

64 * sin(20) = F + 0.3 * 64 * cos(20)

minimum force required to prevent sled from slipping down F = 3.847 N

let the minimum force required to bring sled to motion be F so

mg * sin(20) +  0.3 * mg * cos(20) = F

64 * sin(20) +  0.3 * 64 * cos(20) = F

F = 39.931 N

minimum force required to bring sled to motion F = 39.931 N

if sled is in constant motion then,

F - mg * sin(20) +  0.17 * mg * cos(20) = ma

since velocity is constant then a = 0 m/s^2 so,

F =  mg * sin(20) +  0.17 * mg * cos(20)

F =  64 * sin(20) +  0.17 * 64 * cos(20)

F = 32.113 N

force required to move block in constant velocity = 32.113 N

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