A block weighing #10 kg# is on a plane with an incline of #pi/6# and friction coefficient of #1/2#. How much force, if any, is necessary to keep the block from sliding down?

1 Answer
Mar 1, 2018

Answer:

#6.65N#

Explanation:

enter image source here Here downward component of weight of the block acting to pull the block down along the plane is #mg sin 30=10(1/2)g=5g#

And,frictional force acting upwards to balance that force is #mu mg cos 30=4.32g#(where, #mu# is the coefficient of friction)

So,net force acting on the object is #5g-4.32g=0.68g=6.65N# downwards along the plane,

So,this much force must be supplied to the block upwards along the plane to balance this extra force.