Question

A rough inclined plane is inclined at 30 degrees 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

Answer:

the length of the chain which hang on vertical side is

$x = L\frac{sin\theta + \mu cos\theta}{1 + sin\theta + \mu cos\theta}$

Explanation:

Let the maximum length that hang on the vertical side is "x"

Now we have

weight of the hanging part = weight of the chain on inclined side + friction force

$\frac{x}{L}Mg = \frac{L - x}{L}Mg sin\theta + \mu\frac{L - x}{L}Mg cos\theta$

now we have

$x = (L - x) sin\theta + \mu (L - x)cos\theta$

so we have

$x(1 + sin\theta + \mu cos\theta) = L(sin\theta + \mu cos\theta)$

so the length of the chain which hang on vertical side is

$x = L\frac{sin\theta + \mu cos\theta}{1 + sin\theta + \mu cos\theta}$

$\theta = 30^o$

#Learn

Topic : Friction force

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