Question

A block weighing 71.5 N rests on a plane inclined at 24.1° to the horizontal. The coefficient of the static and kinetic frictions are 0.26 and 0.13 respectively. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the block from slipping?
B)What is the minimum magnitude of F that will start the block moving up the plane?
C)What is the magnitude of F is required to move the block up the plane at constant velocity?

Answer 1

A. F = Fp = 71.5*sin24.1 = 29.20 N.

B. Fn = 71.5*cos24.1 = 65.27 N. = Normal
force = Force perpendicular to the plane

Fs = u*Fn = 0.26 * 65.27 = 16.97 N =
Force of static friction.

F-Fp-Fs = m*a = m*0 = 0
F-29.20-16.97 = 0
F = 46.17 N.

C. Fk = u*Fn = 0.13 * 65.27 = 8.49 N. =
Force of kinetic friction.

F-Fp-Fk = m*a = m*0 = 0
F-29.20-8.49 = 0
F = 37.69 N.