Question

Consider an inclined plane whose upper half is
rough while lower half is smooth. Friction
coefficient of the rough surface is 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?
rough
smooth
37°
lo
3
@
al
string of land​

Answer 1

The fraction of rod required is M.tanФ/(0.8 + tanФ).

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)g.cosФ

⇒ 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)

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