physics
posted by Taylor on .
A loaded penguin sled weighing 75 N rests on a plane inclined at angle θ = 24° to the horizontal (Fig. 628). 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?

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. 
part a is wrong

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