Calculate the clockwise moment about $O$ .

50 g f

c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5000 g f

c m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2500 g f

c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

25 g f

c m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $5000 g f$ $c m$
The turning effect of a force is known as the moment. It is the product of the force multiplied by the perpendicular distance from the line of action of the force to the pivot or point where the object will turn. That is, **Moment = Force * distance.
If the moment of a force turns or rotates the body in clockwise direction, then it is called clockwise moment.
In this case, the weight of 250 gf at the distance of 20 cm from the center point O, rotates the ruler in the clockwise direction.
Therefore, the clockwise moment is given as $C l o c k w i s e m o m e n t = F o r c e \times a r m d i s t a n c e$ .
That is, $C l o c k w i s e m o m e n t = 250 g f \times 20 c m = 5000 g f - c m$ .
Hence, the anticlockwise moment in this case is 5000 gf-cm.**

Suggest Corrections

Similar questions

The diagram in the figure shows a uniform metre rule weighing 100 gf, pivoted at its centre O. Two weights 150 gf and 250 gf hang from the points A and B of the metre rule such that OA = 40 cm and OB = 20 cm. Calculate:

(i) the total anticlockwise moment about O,

(ii) the total clockwise moment about O,

(iii) the difference of anticlockwise and clockwise moments,

and (iv) the distance from O where a 100 gf weight should be placed to balance the metre rule.

The moment of force due to a force of $10 N$ acting about a point O is $5 N m$ . Calculate the distance of the point of application of the force from the point O.