Physics [5 Mark Question Answer]

Take a triangular lamina. Make three fine holes at a, b, c near the edge of the triangular lamina. Now suspend the given lamina along with a plumb line from hole 'a'. Check that the lamina is free to oscillate about the point of suspension. When lamina has come to rest, draw straight line ad along the plumb line. Repeat the experiment by suspending the lamina through hole 'b' and then through hole 'c' for which we get straight lines to be and cf respectively. It is noticed that the lines ad, be and cf intersect each other at a common point G which is the position of the centre of gravity of triangular lamina i.e. the point of intersection of medians.

Perpendicular distance of point A from the force F = 10 N at B is 0.5m, while it is zero from the force F = 10 N at A

Hence moment of force about A is

= 10 N x 0.5m = 5 Nm(clockwise)

Perpendicular distance of the point B from the Force F = 10 N at A is 0.5m while it is zero from the force F = 10 N at B.

Hence moment of force about B is

= 10N x 0.5m = 5Nm (clockwise)

Perpendicular distance of point O from either of the forces F = 10N is 0.25 m

The moment if force F (=10N) at A about O = 10N x 0.25m

= 2.5 Nm (clockwise)

And a moment of force F (=10N) at B about O

= 10 N x 0.25m = 2.5 Nm(clockwise)

Hence total moment of the two forces about O

= 2.5 + 2.5 = 5 Nm (clockwise)

Anticlockwise moment = 40 gf x (50 - 10) cm

= 40 gf x 40 cm = 1600 gf x cm

The clockwise moment is not equal to the clockwise moment. Hence the metre rule is not in equilibrium and it will turn anticlockwise.

When the body is pivoted at a point, the force applied on the body at a suitable point rotates the body about the axis passing through the pivoted point.
The direction of rotation can be changed by changing the point of application of force. The given figure shows the anticlockwise and clockwise moment s produced in a disc pivoted at its centre by changing the point of application of force F from A to B

At A and B, two equal and opposite forces each of magnitude F are applied. The two forces rotate the bar in the anticlockwise direction. The perpendicular distance between two forces is AB which is called couple arm.

Moment of Force F at the end A

= F x OA (anticlockwise)
Moment of force F at the end B
= F x OB (anticlockwise)
Total moment of couple = F x OA + F x OB
= F x (OA + OB) = F x AB
= F x d (anticlockwise)
= Either force x perpendicular distance between the two forces (or couple arm)
Thus, Moment of couple = Force x Couple arm