A crate of potatoes of mass 17.2 kg is on a ramp with angle of incline 30° to the horizontal. The coefficients?
A crate of potatoes of mass 17.2 kg is on a ramp with angle of incline 30° to the horizontal. The coefficients of friction are μs = 0.75 and μk = 0.40. Find the normal force (magnitude) on the crate if the crate is at rest.
? N
Find the frictional force (magnitude and direction) on the crate if the crate is at rest.
magnitude ? N
direction ?
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Mass of the potato crate = M = 17.2 kg
Weight of the potato crate = W = Mg = 17.2*9.8 N acting vertically downwards.
Angle of inclination of the ramp = Θ = 30º
Resolving W in to components parallel and perpendicular to the incline, we have
(i) Force parallel to the incline downwards = P = W sin Θ
= (17.2*9.8) sin 30 = 17.2*9.8*0.5 N = 84.28 N
(ii) Force perpendicular to the incline downwards = N = W cos Θ
= (17.2*9.8) cos 30 = 17.2*9.8*√3/2 = 145.973 N
=> Normal force on the crate at rest = N = 145.973 N (upward) -------- (1)
Force of static(limiting) friction = µsN = f(max) = µsW cos Θ
= 0.75*145.973 N = 109.48 N
where f(max) = maximum value of the frictional force.
P < µsW cos Θ
=> P < f(max)
=> the crate is at rest
=> P - f = 0, where f = frictional force acting parallel to the incline (upward)
=> f = P = W sin Θ = 84.28 N --------- (2)
Friction is a self adjusting force.
Force along the inclined plane due to weight of potatoes=17.2*g*sin30
=172/2=86N
Force of friction (maximum)= 0.75*17.2*g*cos30
=111.714
As this value is greater than required for it to move it will remain at rest.
Friction will self adjust to 86 N.
Normal will remain as 17.2*g*cos30=148.95N
Direction of friction is to oppose relative motion, so it would be up on plane.