Find the coefficient of friction between the particle and the plane in the following case.A particle of mass 0.6 kg is held in equilibrium on a...

For the particle at equilibrium on the inclined plane in
the problem, there are three forces acting on it. One of them is the gravitational pull
of the Earth which acts on the particle in a downward direction. The other is a
horizontal force that is exerted on the particle pushing it towards the plane. These
result in a third force due to friction which is in the opposite direction to that in
which the particle can move.

The frictional force is given
as Fc*N, where Fc is the coefficient of kinetic friction and N is the normal
force.

Here N is the sum of the forces normal to the plane
due to a component of the gravitational force and a component of the horizontal force
being applied.

N = 0.6*9.8*sin 60 + 5*sin
30

The force acting downwards parallel to the plane is
(5*cos 30 - 0.6*9.8*cos 60)

As the force acting downward is
equal to the force acting upward.

Fc* N = (5*cos 30 -
0.6*9.8 cos 60)

=> Fc * ( 0.6*9.8*sin 60 + 5*sin 30)
= (5*cos 30 - 0.6*9.8*cos 60)

=> Fc = (5*cos 30 -
0.6*9.8*cos 60) / ( 0.6*9.8*sin 60 + 5*sin 30)

=> Fc
= 0.1830

Therefore the coefficient of
friction is 0.1830.