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?

Answer 1

Wt. of block = 71.5 N.

Fp = 71.5*sin24.1 = 29.20 N. = Force
parallel to the plane.

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.