A uniform 1.3-kg rod that is 0.67 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both springs hang straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are 58 N/m and 36 N/m. Find the angle that the rod makes with the horizontal.

α = 5.75°

Explanation:

In this case, the problem states that both springs have identical lenghts and we also have theri constant. We want to know the angle of the rod with the horizontal. This can be found with the following expression:

sinα = Δx/L

α = sin⁻¹ (Δx/L)    (1)

However, we do not have Δx. This can be found when half of the weight of the rod is balanced. In this way:

F₁ = k₁*x₁   > x₁ = F₁ / k₁   (2)

And the force is the weight in half so: F₁ = mg/2

Replacing in (2) we have:

x₁ = (1.3 * 9.8) / (2 * 58) = 0.1098 m

Doing the same thing with the other spring, we have:

x₂ = (1.3 * 9.8) / (2 * 36) = 0.1769 m

Now the difference will be Δx:

Δx = 0.1769 - 0.1098 = 0.0671 m

Finally, we can calculate the angle α, from (1):

α = sin⁻¹(0.0671 / 0.67)

Hope this helps

i believe the answer is a