Question

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The diagram below shows a uniform bar supported at a middle point of a weight of 40 g f is placed at a distance of 40 cm to the left of the point how can use balance the bar with the weight of 80gf

•FOR DIAGRAM REFER THE ATTACHMENT


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Answer 1

${\huge{\bold{\underline{Answer:-}}}}$

Anticlockwise moment = 40gf × 40 cm

Clockwise moment = 80 gf × d cm

From the principle of moments,

Anticlockwise moment = Clockwise moment

40 gf × 40 cm = 80 gf × d

So, d = 40gf x 40/80 = 20cm to the righte of point O.

$\huge\tt\colorbox{orange}{Hope it helps u}$

Answer 2

Given:

The diagram below shows a uniform bar supported at a middle point of a weight of 40 gF is placed at a distance of 40 cm to the left of the point.

To find:

Where should a weight of 80 gF be placed in order to balance the scale?

Calculation:

Since we will be considering ROTATIONAL EQUILIBRIUM, the net torque acting on the scale (w.r.t axis at point O) will be zero.

• Let left side lengths be negative and vice-versa.

$\therefore \: \sum(\tau) = 0$

$\implies \: \{ 40 \times ( - 40) \}+ \{80 \times (r) \} = 0$

$\implies \: - 1600+ 80r = 0$

$\implies \: 80r = 1600$

$\implies \: r = 20 \: cm$

Since value of r is positive, the 80 gF weight is to placed on 20 cm mark on right side of scale.