Question

A uniform 1.5-kg rod that is 0.80 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 56 N/m and 35 N/m. Find the angle that the rod makes with the horizontal?

Answer 1

Answer:

θ=5.65°

Explanation:

Given Data

Mass m=1.5 kg

Length L=0.80 m

First spring constant k₁=35 N/m

Second spring constant k₂=56 N/m

To find

Angle θ

Solution

As the both springs take half load so apply Hooks Law:

Force= Spring Constant ×Spring stretch

F=kx

x=F/k

as

$d=x_{1}-x_{2}\\ as \\x=F/k\\so\\d=\frac{F_{1} }{k_{1}} -\frac{F_{2}}{k_{2}}\\ Where \\F=1/2mg\\d=\frac{(1/2)mg}{k_{1}} -\frac{(1/2)mg}{k_{2}}\\ d=\frac{mg}{2}(\frac{1}{k_{1}} -\frac{1}{k_{2}} )\\ And\\Sin\alpha=d/L\\\\alpha =sin^{-1}[\frac{mg}{2L}(1/k_{1}-1/k_{2})]\\\alpha =sin^{-1}[\frac{(1.5kg)(9.8m/s^{2} )}{2(0.80m)}(1/35Nm-1/56Nm) ]\\\alpha =5.65^{o}$

θ=5.65°