A loaded penguin sled weighing 68.0 N rests on a plane inclined at angle theta =

Question

A loaded penguin sled weighing 68.0 N rests on a plane inclined at angle theta = 22.0 degree to the horizontal (see the figure). Between the sled and the plane, the coefficient of static friction is 0.230, and the coefficient of kinetic friction is 0.130. (a) What is the minimum magnitude of the force F vector, 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

here,

weight of sled , W = 68 N

a)

the minimum force required be F

equaitng the forces

F + us * W * cos(theta) = W * sin(theta)

F = 68 * ( sin(22) - 0.23 * cos(22))

F = 11 N

b)

for the block to move up the plane

let the force required be F

equaitng the forces

F - us * W * cos(theta) = W * sin(theta)

F = 68 * ( sin(22) + 0.23 * cos(22))

F = 40 N

c)

for the block to move at constant velocity

the coefficient of kinetic friction is used

let the force required be F

equaitng the forces

F - us * W * cos(theta) = W * sin(theta)

F = 68 * ( sin(22) + 0.13 * cos(22))

F = 33.7 N

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