Question

A 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

A 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.

To find : The position of point of balance.

solution : let A and B are ends of given rods where 10N and 15 N weights are suspended respectively.

let x distance from end A of rod is balancing point.

we know,

at balancing point,

net torque = 0

⇒10N × x (←) + 6N × (1/2 - x) (→) + 15N × (1 - x) (→) = 0

⇒10x - 6(1/2 - x) - 15(1 - x) = 0

⇒10x - 3 + 6x - 15 + 15x = 0

⇒31x = 18

⇒x = 18/31 m ≈ 58 cm

Therefore the position of balancing point is 58 cm from end of weight 10N

Answer 2

Answer:

       Given data    W1  = 6 N            W2 = 10 N         W3 = 15 N     R = 1m         r2 =1- x    r3 = x   r1 = 0.5-x

 

Solution  

Using principle of moment  

                                                  torque clock = torque anti

                                                W3 × r3 = W1 × r1 + W2 × r2

                                                15 x = 6( 0.5- x) + 10(1- x)

                                                15 x = 3 – 6 x + 10 -10 x

                                                15 x + 6 x + 10 x = 13

                                                31 x = 13

                                                   X = 13 / 31 = 0.42 m  

Explanation: