Question

A uniform 1.8-kg rod that is 0.88 m long is suspended at rest from the ceiling by two springs, one at each end. Both springs hang straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are 36 N/m and 52 N/m. Find the angle that the rod makes with the horizontal.

Answer 1

Given,

Mass of the rod, m = 1.8 kg

Length L = 0.88 m

The spring constants, k₁ = 52 N/m and k₂ = 36 N/m

As the rod is in equilibrium

The net torque acting on the rod is zero

On the right side of the rod,

m*g*L/2 = F*L

where, F is the restoring force

m*g / 2 = k₁*x₁

1.8 * 9.8 / 2 = 52 * x₁

x₁ = 0.169 m

On the left side of the rod

m*g*L/2 = k₂*x₂*L

1.8 * 9.8 / 2 = 36 * x₂

x₂ = 0.245 m

We know, dx = x₂ - x₁

   = 0.245 - 0.169

= 0.076 m

sin = dx / L

sin = 0.076 / 0.88

sin = 0.0863

= sin^-1 (0.0863)

= 4.950 ^o

answered by: anonymous