Question 13Figure shows a uniform metre rule placed on a fulcrum at its mid-point O and having a weight 40gf at the 10 cm mark and a weight of 20 gf at the 90 cm mark.(i) Is the metre rule in equilibrium? If not, how will the rule turn?(ii) How can the rule be brought in equilibrium by using an additional weight of 40gf?

Question

Accepted Answer

(i) We know that when anticlockwise moment is equal to clockwise moment then the rule will be in equilibrium.

Let us check,

(a) Anticlockwise moment = Fr

$40gf \times (50-10)cm = 40 \times 40 \\[0.5em] = 1600 gfcm$

(b) Clockwise moment = Fr

$20gf \times (90-50)cm = 20 \times 40 \\[0.5em] = 800gfcm$

So, the meter rule will not be in equilibrium. It will turn in anticlockwise direction.

(ii) To balance the rule in equilibrium 40gf should be kept on right side.

Let distance from middle = d cm

So clockwise moment becomes

$800 gf cm + (40gf \times d cm)$

As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment

$1600gfcm = 800gfcm + 40gf \times d cm$

Therefore,

$d = \dfrac{800}{40} = 20cm$

So, the additional weight should be placed at 70cm mark to bring the rule in equilibrium.

Force — Question 13

Question 13