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...

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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.**