Question

A uniform 1.6-kg rod that is 0.89 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 31 N/m and 63 N/m. Find the angle that the rod makes with the horizontal.

Answer 1

Answer:

8.27°

Explanation:

To angle difference will be determined by the difference in the displacement of the springs, produced by the weight of the center of mass of the rod.

$d=y_1-y_2=\frac{F_1}{k_1}-\frac{F_2}{k_2}=\frac{0.5mg}{31N/m}-\frac{0.5mg}{63N/m}\\\\d=0.5(1.6kg)(9.8m/s^2)[\frac{1}{31N/m}-\frac{1}{63N/m}]=0.128m$

by a simple trigonometric relation you obtain that the angle:

$sin\theta=\frac{d}{l}=\frac{0.128m}{0.89m}=0.144\\\\\theta=sin^{-1}(0.144)=8.27\°$

hence, the angle between the rod and the horizontal is 8.27°