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

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.