Question

A rough inclined plane is inclined at 300 to the horizontal. A uniform chain of length L is partly on the inclined plane and partly hanging from the top of the incline. If the coefficient of friction of μ b/w the chain and inclined plane, the maximum length of the hanging part to prevent the chain from falling vertically is

Answer 1

(1 + $\sqrt{3}$μ)L/$\sqrt{3}$(μ  + $\sqrt{3}$)

Explanation:

Given L is the overall length of the chain

Let the hanging part be x

so,

L → M

x → $\frac{x}{L} M$

Altitude = L - x

L - x = $\frac{( L - x)}{L} M$

A.T.Q.

Two blocks, one on vertical(Mx) and horizontal sides(ML - x) each.

So,

Tension(T) = Mxg

T = (x/L)Mg ...(i)

Now, it must be prevented to fall vertically,

M * (L - x)g sin θ + frictional force value

= M(L−x) g[sinθ + cosθ]

Now by solving,

x/L Mg = [(L−x)/L] Mg [sin30° +  μcos30°]

x = (L - x)1/2 + μ(L−x)$\sqrt{3}$/2

∵ x = (1 + $\sqrt{3}$μ)L/$\sqrt{3}$(μ  + $\sqrt{3}$)

Learn more: Friction

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