The wooden door of a building is $3 m$ broad. It can be opened by applying a force of $100 N$ at the middle of the door. Calculate the least force which can open the door.

$55 N$

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$52 N$

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$60 N$

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$50 N$

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Solution

The correct option is D
$50 N$

Distance between middle of the gate and fixed end of the gate = $\frac{3}{2} = 1.5 m$
Moment of Force about the fixed edge:
= $F \times d = 100 \times 1.5 = 150 N m$
The least force ( $F_{1}$ ) can be applied at the free end of the door.
$∴$ By principal of moments:
Then, the moment of force = $F_{1} \times 3$
$∴ 150 = F_{1} \times 3$
$∴ F_{1} = 50 N$

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