Question

Fig.  shows a uniform meter scale weighing 100 N pivoted at its centre. Two weights of 500 N and 300 N are hung from the ruler as shown in fig.

(i)    Calculate total clockwise and anticlockwise moments.
(ii)    Calculate difference in clockwise moment and anticlockwise moment.
(iii)    Calculate the distance from O where a 100 N weight should be suspended to balance the meter scale.

Answer 1

Let the weight 500 N be hung from the 40cm mark and the weight 300N be hung from the 80 cm mark

(i) Then, total clockwise moment = 500 x (50 - 40) = 500 x 10 = 5000 Nm

Total anticlockwise moment = 300 x (80 - 50) = 300 x 30 = 9000 Nm

(ii) Difference in clockwise and anticlockwise moment = 9000 - 5000  = 4000Nm

(iii) Let the 100 N weight be hung at a distance 'd' from the point 'o' to its left

Then,total clockwise moment = 5000 + 100 d

In balanced condition, sum of clockwise moments = sum of anticlockwise moments 

5000 + 100 d = 9000

or 100 d = 4000

or d = 40cm

the weight 100 N should be a hung from the 40cm mark so as to balance the scale.