A man is standing on a plank of weight $100 k g f$ pivoted at A and B as shown in figure. If the reaction force by pivot B is $66 k g f$ , find the weight of the man.

50 k g f

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60 k g f

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70 k g f

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80 k g f

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Solution

The correct option is D $80 k g f$
Let the weight of the man be W kgf.
Let reaction force at pivot A and B is $R_{A}$ and $R_{B}$ respectively.
Since pivot A is at equilibrium,
Clockwise moment = anticlockwise moment
$(W \times 2 m) + (100 k g f \times 5 m) = (66 k g f \times 10 m)$
$\Rightarrow W = 80 k g f$ .
Thus, the weight of the man is $80 k g f$ .

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A man is standing on a plank of weight 100 kgf pivoted at A and B as shown in figure. If the reaction force by pivot B is 66 kgf, find the weight of the man.

The correct option is D 80 kgfLet the weight of the man be W kgf. Let reaction force at pivot A and B is RA and RB respectively. Since pivot A is at equilibrium, Clockwise moment = anticlockwise moment (W×2 m)+(100 kgf×5 m)=(66 kgf×10 m) ⇒W=80 kgf. Thus, the weight of the man is 80 kgf.

A man is standing on a plank of weight $100 k g f$ pivoted at A and B as shown in figure. If the reaction force by pivot B is $66 k g f$ , find the weight of the man.