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A uniform 1.2-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 65 N/m and 30 N/m. Find the angle that the rod makes with the horizontal.


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The angle that the rod makes with the horizontal is 9⁰.

What is the vertical direction for left spring?

The vertical direction of the left spring is determined  by applying the principle of equilibrium force as shown below.

The given parameters;

  • mass of the rod, m = 1.2 kg
  • weight of the rod, w = mg = 1.2 x 9.8 = 11.76 N
  • spring constant for left spring, K_L = 65 N/m
  • spring constant for right spring, K_r = 30 N/m
  • stretch in the left spring, = X_L
  • stretch in the right spring, = X_r
  • length of the rod, L  = 0.67 m
  • angle the rod makes with the horizontal, θ = ?

Apply the principle of equilibrium of force in vertical direction for left spring;

K_L(X_L) = 0.5(W)

65(X_L) = 0.5(11.76)

X_L = 5.88/65

X_L = 0.09 m

Apply the principle of equilibrium of force in vertical direction for right spring;

K_r(X_r) = 0.5(W)

30(X_r) = 0.5(11.76)

X_r = 5.88/30

X_r = 0.196 m

The angle made with the horizontal is calculated as;

θ = arc tan(X_r - X_L) / L

θ = arc tan[(0.196 - 0.09) / 0.67]

θ = arc tan(0.158)

θ = 8.99⁰

θ ≈ 9⁰

Learn more about angle made by rod here: https://brainly.com/question/14085205

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