Asked by Ameen
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.
Answers
Answered by
Henry
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.
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.
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