Question

Question #996b0

Answer 1

If the springs are attached asymmetrically with respect to the center of gravity of the weight in a very specific way, then yes.

Explanation:

If the weight `W` is to stay in equilibrium, the net force and the net torque on it must both vanish.

If the elongation `x` is the same in both the springs (spring constants `k_1` and `k_2`, respectively), then they will exert upward forces of `k_1 x` and `k_2x`, respectively. If the center of gravity of the weight is at distances of `d_1` and `d_2` from the points where the two springs are attached to it. we have

(Force balance) `(k_1+k_2)x = W`
(Torque balance) `k_1x times d_1 = k_2 x times d_2`

So, we must have

`d_1/d_2 = k_2/k_1`

for this to happen.