Chapter 12, Problem 58P

(a)

To determine

To Find:Coefficient of friction between the beam and the wall.

Answer to Problem 58P

  $0.75$

Explanation of Solution

Given data:

Weight of steel beam $w=22\text{lb/ft}$

  $\begin{array}{l}\text{=}\left(\text{22}\frac{\text{lb}}{\text{ft}}\right)\left(\frac{\text{1}\text{kg}}{\text{2}\text{.205}\text{lb}}\right)\left(\frac{\text{3}\text{.281}\text{ft}}{\text{m}}\right)\\ \text{=32}\text{.735}\text{kg/m}\end{array}$

Length of the beam, $L=4.0\text{m}$

Length of the cable, $x=5.0\text{m}$

Formula used:

For translational and rotational equilibrium:

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

Calculation:

  

Horizontal component of net force

  $\begin{array}{l}{\displaystyle \sum {\overline{F}}_x}=0\\ F_n-T\cos \theta =0\mathrm{.......}\left(1\right)\\ F_n=T\cos \theta \end{array}$

Vertical component of net force

  $\begin{array}{l}{\displaystyle \sum {\overline{F}}_y}=0\\ f_s+T\sin \theta -m_{beam}g=0\\ {\mu }_s{F}_n=m_{beam}g-T\sin \theta \left(\text{As }{f}_s={\mu }_s{F}_n\right)\mathrm{......}\left(2\right)\\ {\mu }_s=\frac{m_{beam}g-T\sin \theta }{F_n}\end{array}$

From $\left(1\right)$ and $\left(2\right)$

  $\begin{array}{c}{\mu }_s=\frac{m_{beam}g-T\sin \theta }{F_n}\\ =\frac{m_{beam}g}{T\cos \theta }-T\tan \theta \end{array}$

  $m_{beam}=\lambda L$

Where, $\lambda$ is the linear density and L is the length of the beam.

  ${\mu }_s=\frac{\lambda Lg}{T\cos \theta }-\tan \theta \mathrm{............}\left(3\right)$

  ${\displaystyle \sum \tau =0}$

  $\begin{array}{l}\left(T\sin \theta \right)L-m_{beam}g\left(L/2\right)=0\\ \left(T\sin \theta \right)L=m_{beam}g\left(L/2\right)\mathrm{...........}\left(4\right)\\ T=\frac{\lambda Lg}{2\sin \theta }\end{array}$

Equation $\left(3\right)$ becomes

  $\begin{array}{c}{\mu }_s=\frac{\lambda Lg}{\frac{\lambda Lg}{2\sin \theta }\cos \theta }-\tan \theta \\ =2\tan \theta -\tan \theta \\ =\tan \theta \\ =\frac{3}{4}\\ =0.75\end{array}$

Conclusion:

Thus, the coefficient of friction between the beam and the wall is $\mu =0.75$ .

(b)

To determine

To Find:Tension in the cable.

Answer to Problem 58P

  $1.1\text{ kN}$

Explanation of Solution

Given data:

  $\begin{array}{l}\lambda =32.735\text{ lb/ft}\\ \text{L=4}\text{.0 m}\\ \text{g = 9}{\text{.8 m/s}}^2\\ \theta ={\tan }^{-1}\left(0.75\right)\end{array}$

Formula used:

From part (a):

  $T=\frac{\lambda Lg}{2\sin \theta }$

Calculation:

Substitute the values and solve:

  $\begin{array}{c}T=\frac{\lambda Lg}{2\sin \theta }\\ =\frac{\left(\text{32}\text{.735}\text{lb/ft}\right)\left(\text{4}\text{.0}\text{m}\right)\left(\text{9}\text{.8}{\text{m/s}}^{\text{2}}\right)}{\text{2sin}\left({\text{tan}}^{\text{-1}}\left(\text{0}\text{.75}\right)\right)}\\ \text{=1069}\text{.3433}\text{N}\left(\frac{\text{1}\text{kN}}{\text{1000}\text{N}}\right)\\ \approx 1.1\text{kN}\end{array}$

Conclusion:

Thus, the tension in the cable is $T=1.1\text{}\text{kN}$ .