weight component down slope = 98.1 sin 25
= 41.5 N
weight component normal to ramp = 98.1 cos 25
= 88.9 N
max static friction = 88.9*.5= 44.5 N
initial weight of bucket = 2*9.81 = 19.6 N
force up ramp = 19.6 N
force down ramp = 41.5 N
Net force before friction down ramp =
41.5 - 19.6 = 21.9 N
that is much less than the static friction so no motion results
Friction = 21.9 N to prevent slip down ramp.
now add water, force up ramp = 6*9.81 = 58.9 N
again force down ramp = 41.5 N
total force up ramp before friction =58.9-41.5 = 17.4 N up
again less than the maximum static friction so no motion
friction force = 17.4 down ramp to prevent motion up ramp
A wooden box with a mass of 10.0kg rests on a ramp that is inclined at angle of 25 degrees to the horizontal. A rope attached to the box runs parallel to the ramp and then passes over a frictionless pulley. A bucket with a mass of m hangs from the end of the rope. The coefficient of static friction between the ramp and the box is 0.50. The coefficient of the kinetic friction between the ramp and the box is 0.35.
suppose the bucket has a mass of 2.0 kg. what is the friction force exerted on the box by the ramp?
suppose water is added to the bucket so that the total mass of the bucket and its contents is 6.0 kg. what is the friction force exerted on the box by the ramp?
1 answer
1 answer