Question

Consider an inclined plane whose upper half is rough while lower half is smooth. Friction coefficient of rough surface u=0.8. A heavy rope of mass M is placed on the inclined plane such that it remains in equilibrium. What minimum fraction of rope is required on rough surface for equilibrium?

Answer 1

Explanation:

Let m is kept in lower half and M is in upper half ;)

Let the angle of inclined be Ф.

There will be downward pull of mgsinФ on the lower half.

This will be balanced by maximum upward pull of friction u(M - m)gcosФ

⇒ mgsinФ = u(M - m)gcosФ

⇒ msinФ = u(M - m)cosФ

⇒ sinФ/cosФ = u(M - m)/m

⇒ tan Ф/u = M/m - 1

⇒ M/m = 1 + tanФ/u

⇒ M/m = (u + tanФ)/u

⇒ m = uM /(u + tan Ф)

This is the fraction of mass required on lower surface.

Upper surface requires : M - uM / (u + tanФ)

⇒ (Mu + MtanФ - uM)/(u + tanФ)

⇒ MtanФ/(u + tanФ)

⇒ MtanФ/(0.8 + tanФ)

#GIRLSOPHILE

Answer 2

The minimum fraction of rope which is required on rough surface for equilibrium is $\frac{F}{7.84\\\times Cos \theta \\ M}$.

Explanation:

We know that the force of friction is

                  ⇒ F = μ x N

where, u is the friction coefficient = 0.8

                 N is the normal = mg Cos θ

  θ is the angle between the plane and the inclination.

 m is the mass of the rope that is required to make it in equilibrium. M is the total mass of the rope.

          g = 9.8 m/$s^{2}$

       ⇒ F = μ x mg Cos θ

           ⇒ m = $\frac{F}{7.84\\\times Cos \theta \\ }$

       minimum fraction $\frac{m}{M}$ = $\frac{F}{7.84\\\times Cos \theta \\ M}$