The given figure shows a weightless lever in equilibrium. Neglect friction at the fulcrum F.

Calculate the mechanical advantage for the given lever.

$\frac{E f f o r t a r m}{L o a d a r m}$

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$\frac{l o a d a r m}{E f f o r t a r m}$

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$\frac{2 \times E f f o r t a r m}{L o a d a r m}$

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$\frac{E f f o r t a r m}{2 \times L o a d a r m}$

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Solution

The correct option is A
$\frac{E f f o r t a r m}{L o a d a r m}$

The mechanical advantage (MA) of machine is the ratio of the resistance to overcome to the effort applied to the machine
$∴$ MA $= \frac{R e s i s t a n c e o v e r c o m e}{E f f o r t a p p l i e d}$
By the principle of moments, we have
Load $\times$ Load arm = Effort $\times$ Effort arm
or L $\times$ b = E $\times$ a, where a and b are the effort arm and the load arm respectively.
or $\frac{L}{E} = \frac{a}{b}$
i.e. MA of a lever $= \frac{E f f o r t a r m}{L o a d a r m}$

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Similar questions

For a lever, $M e c h a n i c a l A d v a n t a g e = \frac{l o a d a r m}{e f f o r t a r m}$ .

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4

90 cm

Calculate the mechanical advantage of a lever in which the effort arm is 50 cm and the load arm is 2 cm.

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