Suppose there is friction (static coefficient = 0.22, and kinetic coefficient = 0.15). With this friction, what force (if any) is necessary to hold the 30kg mass stationary on a 35 degree incline?

Answer 1

Wm = m*g = 30kg * 9.8N/kg = 294 N. = Wt.
of the mass.

Fp = 294*sin35 = 168.6 N. = Force parallel to the incline.

Fn = 294*cos35 = 240.8 N. = Normal force
or force perpendicular to the incline.

Fs = u*Fn = 0.22 * 240.8 = 52.98 N. =
Force of static friction.

Fap-Fp-Fs = 0
Fap = Fp+Fs = 168.6 + 52.98 = 221.6 N. =
Force applied.