A loaded penguin sled weighing 75 N rests on a plane inclined at angle θ = 24° to the horizontal (Fig. 6-28). Between the sled and the plane, the coefficient of static friction is 0.25, and the coefficient of kinetic friction is 0.13. (a) What is the minimum magnitude of the force , 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?

Answer 1

Fs = (75N,24deg).
Fp=75sin24 = 30.51N. = Force parallel to plane acting down the plane.
Fv = 75cos24 = 68.52N. = Force perpen-
dicular to plane.

Ff = u*Fv = 0.25 * 68.52 = 17.13N. =
Force of static friction.
Fk = 0.13 * 68.52 = 8.91 = Force of
kinetic friction.

a. Fap - Fp - Fk = 0,
Fap=Fp + Ff = 30.51 + 8.91 = 39.42N =
Min Force applied.

b. Fap - Fp - Ff = 0,
Fap = Fp + Ff = 30.51 + 17.13 = 47.64N.
= Min force applied.

c. Fap = Fp + Fk = 30.51 + 8.91 = 39.42N.

Answer 2

part a is wrong

Answer 3

pt A: Fp - Ff. in this case, you would do 30.51 - 17.13 = 13.38