A crate weighing 400N rests on aplane inclined at an angle of 20 degrees to the horizontal. there is a force of friction of 120N acting up the ramp. Determine (i) the resultant force acting on the crate, (ii) the contact,or normal force between the ramp and the crate.

Answer 1

Wc = 400 N. = Wt. of crate.

Fc = 400N @ 20o. = Force of crate.
Fp = 400*sin20 = 136.8 N. = Force parallel to incline.
Fv = 400*cos20 = 376 N. = Normal =
Force perpendicular to incline.

Fr = Fp-Ff = 136.8-120 = 16.8 N. = Resultant force.