Question

uniform rod 1 m long with weight 6 N can be supported in a horizontal
position on a sharp edge with weights of 10 N and 15 N suspended from
its ends. What is the position of point of balance?​

Answer 1

Answer:

Explanation:

The weight will act from the middle point downwards . The balancing point will be shifted from middle point by distance x towards the heavy weight that is 15 N .

Taking torque about the balancing point

6 x + 10 ( .5 + x ) = 15 ( .5 - x )

6x + 5 + 10x = 7.5 - 15 x

31 x = 2.5

x = .08 m

The balance point will be at .08 m away from middle point of the rod towards the force of 15 N .