Question

a uniform metre rule is pivoted at its centre and weights of 5 N and 12 N are hung at the 3 cm and 5 cm marks respectively. how far from the pivot must a 25 N weight be hung to balance the metre rule horizontally? ​

Answer 1

In order to balance the stick on the pivot, the total "moments" must be equal on both sides. A "moment" is (a weight) x (its distance from the center).

for the 5N weight: Moment = (5N) x (3 cm) = 15 N-cm

for the 12N weight: Moment = (12N) x (5 cm) = 60 N-cm

Sum of the moments trying to pull the stick down on that side = 75 N-cm

Whatever we hang on the other side has to provide a moment of 75 N-cm in the other direction. We have a 25N weight. Where should we hang it ?

(25N) x (distance from the pivot) = 75 N-cm

Distance from the pivot = (75 N-cm) / (25 N)

Distance from the pivot = 3 cm

Answer 2

The distance do a 25N be hung from the pivot is 31.2cm