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
[tex]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}[/tex]
θ=5.65°
