Question

A rod is 20 cm long. Its 8 kg and 12 kg weights are hanging on both ends. If the weight of the rod is 6 kg, then how far will the balance point be from the weight of 12 kg

Answer 1

Explanation:

solve the abv eqn for x. hope you get the solution

Answer 2

Let the balance point of the rod be x (in cm) from the weight of 12 kg.

Here the weights at the ends of the rod are 8 kgwt and 12 kgwt.

Since the length of the rod (assumed to be uniform) is 20 cm, the weight of the rod alone is concentrated at a point 10 cm from any of the end, and the weight is 6 kgwt.

Consider the center of weight which is x centimetres away from the point where 12 kgwt is hung. So, from this point,

• 12 kgwt is x centimetres away towards left.
• 6 kgwt (weight of the rod) is (10 - x) centimetres away towards right (since the distance between 12 kgwt and 6 kgwt is 10 cm).
• 8 kgwt is (20 - x) centimetres away towards right (since the length of rod is 20 cm).

We know that, at the center of weight of any body, the body is in equilibrium so that the net torque acting on the body is a null vector. Thus we have the equation (left side is considered as positive),

12x - 6(10 - x) - 8(20 - x) = 0

12x - 60 + 6x - 160 + 8x = 0

26x = 220

x = 220 / 26

x = 8.46 cm

Hence the center of weight is nearly 8.46 cm from the point where 12 kgwt is hung.