Chapter 14, Problem 38PQ

At a museum, a 1300-kg model aircraft is hung from a lightweight beam of length 12.0 m that is free to pivot about its base and is supported by a massless cable (Fig. P14.38). Ignore the mass of the beam. a. What is the tension in the section of the cable between the beam and the wall? b. What are the horizontal and vertical forces that the pivot exerts on the beam?

FIGURE P14.38

(a) From the free-body diagram, the angle that the string tension makes with the beam is θ = 55.0° + 18.0° = 73.0°, and the perpendicular component of the string tension is F_T sin73.0°. Summing torques around the base of the rod gives (Eq. 14.2):

$\begin{array}{l}{\displaystyle \sum \tau =0:-(12.0\text{ m})(1300\text{ kg})(9.81{\text{ m/s}}^2)\cos 55.0°+F_T(12.0\text{ m})\sin 73.0°=0}\\ F_T=\frac{(12.0\text{ m})(1300\text{ kg})(9.81{\text{ m/s}}^2)\cos 55.0°}{(12.0\text{ m})\sin 73.0°}\\ F_T=\boxed{7.65\times {10}^3\text{N}}\end{array}$

Figure P14.38ANS

(b) Using force balance (Eq. 14.1):

$\begin{array}{l}{\displaystyle \sum F_x=0:F_H-F_T\cos 18.0°=0}\\ F_H=F_T\cos 18.0°=\left[\frac{(12.0\text{ m})(1300\text{ kg})(9.81{\text{ m/s}}^2)\cos 55.0°}{(12.0\text{ m})\sin 73.0°}\right]\cos 18.0°=\boxed{7.27\times {10}^3\text{ N}}\\ {\displaystyle \sum F_y=0:F_V-F_T\sin 18.0°-(1300\text{ kg})(9.81{\text{ m/s}}^2)=0}\end{array}$

$\begin{array}{l}{F}_V=F_T\sin 18.0°+(1300\text{ kg})\text{g}\\ F_V=\left[\frac{(12.0\text{ m})(1300\text{ kg})(9.81{\text{ m/s}}^2)\cos 55.0°}{(12.0\text{ m})\sin 73.0°}\right]\sin 18.0°+(1300\text{ kg})(9.81{\text{ m/s}}^2)\\ F_V=\boxed{1.51\times {10}^4\text{N}}\end{array}$