Chapter 12, Problem 59P

To determine

Mass of the object suspended from left end

Answer to Problem 59P

  $1.8\text{ kg}$

Explanation of Solution

Given data:

Length of the rigid beam, $L=2.5\text{m}$

Spring constant, $\text{k =1250 N/m}$

At equilibrium, beam makes an angle $\theta \text{ = 17}{\text{.5}}^{\text{o}}$ with the horizontal

Height of post $\text{=1}\text{.25}\text{m}$

Formula Used:

For translational and rotational equilibrium:

  $\begin{array}{l}{\displaystyle \sum F=0}\\ {\displaystyle \sum \tau =0}\end{array}$

Calculation:

  

  

  ${\displaystyle \sum \tau =0}$

  $mg\left(\frac{1}{2}L\cos \theta \right)-F_{spring}\left(\frac{1}{2}L\sin \phi \right)=0\mathrm{.....}\left(1\right)$

From Hooke’s law

  $F_{spring}=k\Delta {\mathcal{l}}_{spring}$

Equation $\left(1\right)$ becomes

  $mg\left(\frac{1}{2}L\cos \theta \right)-\left(k\Delta {\mathcal{l}}_{spring}\right)\left(\frac{1}{2}L\sin \phi \right)=0\mathrm{.....}\left(2\right)$

From figure $\left(b\right)$ the relation between $\phi \text{ and }\theta$ is:

  $\begin{array}{c}\phi ={\tan }^{-1}\left(\frac{\left(L-L\cos \theta \right)/2}{\left(L+L\sin \theta \right)/2}\right)\\ ={\tan }^{-1}\left(\frac{1-\cos \theta }{1+\sin \theta }\right)\end{array}$

Substitute the value of $\phi$ in $\left(2\right)$ :

  $mg\left(\frac{1}{2}L\cos \theta \right)-\left(k\Delta {\mathcal{l}}_{spring}\right)\left(\frac{1}{2}L\sin \left({\tan }^{-1}\left(\frac{1-\cos \theta }{1+\sin \theta }\right)\right)\right)=0\mathrm{.....}\left(2\right)$

  $\begin{array}{c}mg\left(\frac{1}{2}L\cos \theta \right)-\left(k\Delta {\mathcal{l}}_{spring}\right)\left(\frac{1}{2}L\sin \left({\tan }^{-1}\left(\frac{1-\cos \theta }{1+\sin \theta }\right)\right)\right)\\ m=\frac{\left(k\Delta {\mathcal{l}}_{spring}\right)\left(\sin \left({\tan }^{-1}\left(\frac{1-\cos \theta }{1+\sin \theta }\right)\right)\right)}{g\cos \theta }\mathrm{..........}\left(3\right)\end{array}$

  $\begin{array}{c}\Delta {\mathcal{l}}_{spring}={\mathcal{l}}_{stretched}-{\mathcal{l}}_{unstretched}\\ ={\mathcal{l}}_{stretched}-\frac{1}{2}L\mathrm{..........}\left(4\right)\end{array}$

From figure:

  $\begin{array}{c}2\alpha +\theta =\pi \\ \alpha =\frac{\pi -\theta }{2}\end{array}$

  $\begin{array}{c}\beta =\alpha +\frac{\pi }{2}\\ =\frac{\pi -\theta }{2}+\frac{\pi }{2}\\ =\pi -\frac{\theta }{2}\end{array}$

Apply the law of cosines to the triangle

  $\begin{array}{c}{\mathcal{l}}_{streatched}^2={\left(\frac{L}{2}\right)}^2+\left(L\sin {\left(\frac{\theta }{2}\right)}^2\right)-2\left(\frac{L}{2}\right)\left(L\sin \frac{\theta }{2}\right)\cos \left(\pi -\frac{\theta }{2}\right)\\ =\frac{L^2}{4}+L^2{\sin }^2\left(\frac{\theta }{2}\right)+\frac{L^2}{2}\left(2\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)\right)\\ =\frac{L^2}{4}+L^2{\sin }^2\left(\frac{\theta }{2}\right)+\frac{L^2}{2}\left(\sin \left(\frac{2\theta }{2}\right)\right)\\ =\frac{L^2}{4}+L^2\left(\frac{1-\cos \theta }{2}\right)+\frac{L^2}{2}\sin \theta \\ {\mathcal{l}}_{streatched}^2=\frac{L^2}{4}\left(3-2\left(\cos \theta -\sin \theta \right)\right)\\ {\mathcal{l}}_{streatched}=\frac{L}{2}\sqrt{\left(3-2\left(\cos \theta -\sin \theta \right)\right)}\end{array}$

From equation $\left(4\right)$

  $\begin{array}{c}\Delta {\mathcal{l}}_{spring}=\frac{L}{2}\sqrt{\left(3-2\left(\cos \theta -\sin \theta \right)\right)}-\frac{1}{2}L\\ =\frac{L}{2}\left(\sqrt{\left(3-2\left(\cos \theta -\sin \theta \right)\right)}-1\right)\end{array}$

From $\left(3\right)$

  $m=\frac{\left(\frac{kL}{1}\left(\sqrt{3-2\left(\cos \theta -\sin \theta \right)}-1\right)\right)\left(\sin \left({\tan }^{-1}\left(\frac{1-\cos \theta }{1+\sin \theta }\right)\right)\right)}{g\cos \theta }$

Substitute the values and solve:

  $\begin{array}{c}m=\frac{\left(\text{1250}\text{N/m}\right)\left(\text{2}\text{.5}\text{m}\right)\left(\sqrt{\left(3-2\left(\cos {17.5}^{\text{o}}-\sin {17.5}^{\text{o}}\right)\right)}-1\right)\left(\sin \left({\tan }^{-1}\left(\frac{1-\cos {17.5}^{\text{o}}}{1+\sin {17.5}^{\text{o}}}\right)\right)\right)}{2\left(\text{9}\text{.8}{\text{m/s}}^{\text{2}}\right)\cos {17.5}^{\text{o}}}\\ =\frac{\left(\text{1250}\text{N/m}\right)\left(\text{2}\text{.5}\text{m}\right)\left(\text{0}\text{.301}\right)\left(\text{0}\text{.0355}\right)}{\text{2}\left(\text{9}\text{.8}{\text{m/s}}^{\text{2}}\right)\text{cos17}{\text{.5}}^{\text{o}}}\\ =1.786\text{kg}\\ =1.\text{8}\text{kg}\end{array}$

Conclusion:

Mass of the object is $1.8\text{ kg}$ .