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Standard VI

Physics

Mechanical Advantage

A man and a c...

Question

A man and a child are holding a uniform rod of length $L$ in the horizontal direction in such a way that one fourth weight is supported by the child. If the child is at one end of the rod then the distance of man from another end will be :

\frac{3 L}{4}

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\frac{L}{4}

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\frac{L}{3}

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\frac{2 L}{3}

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Solution

The correct option is C $\frac{L}{3}$
balancing the force $F = m g - \frac{m g}{4} = \frac{3 m g}{4}$
Balancing torque about A
$m g * \frac{l}{2} = \frac{3 m g}{4} (l - x) \Rightarrow 2 l = 3 (l - x) \Rightarrow x = \frac{l}{3}$

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A man and a child are holding a uniform rod of length L in the horizontal direction in such a way that one fourth weight is supported by the child. If the child is at one end of the rod then the distance of man from another end will be :

The correct option is C L3balancing the force F=mg−mg4=3mg4Balancing torque about Amg∗l2=3mg4(l−x)⇒2l=3(l−x)⇒x=l3

A man and a child are holding a uniform rod of length $L$ in the horizontal direction in such a way that one fourth weight is supported by the child. If the child is at one end of the rod then the distance of man from another end will be :

The correct option is C $\frac{L}{3}$
balancing the force $F = m g - \frac{m g}{4} = \frac{3 m g}{4}$
Balancing torque about A
$m g * \frac{l}{2} = \frac{3 m g}{4} (l - x) \Rightarrow 2 l = 3 (l - x) \Rightarrow x = \frac{l}{3}$