Auniform 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?
Answers: 3
![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}](/tpl/images/0470/1329/9c820.png)
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