Figure 12-49ashows a vertical uniform beam of length Lthat is hinged at its lower end. A horizontal force$\overrightarrow{{\mathbf{F}}_{\mathbf{a}}}$ is applied tothe beam at distance yfrom the lower end. The beam remainsvertical because of a cable attached at the upper end, at angle$\mathbf{\theta }$with the horizontal. Figure12-49agives the tension Tin the cableas a function of the position of the applied force given as a fraction y/Lof the beam length. The scale of the Taxis is set by${\mathbf{T}}_{\mathbf{s}}\mathbf{=}\mathbf{600}\mathbf{}\mathbf{N}$.Figuregives the magnitude F_hof the horizontal force on thebeam from the hinge, also as a function of y/L. Evaluate (a) angle$\mathbf{\theta }$and (b) the magnitude of $\overrightarrow{{\mathbf{F}}_{\mathbf{a}}}$.

The scale of the Taxis is set by ${\mathrm{T}}_{\mathrm{s}}=600\mathrm{N}$.

If net torque acting on the system is zero, the object is said to be in equilibrium. This can happen when there are more than one torque acting on the object in a different direction such that the vector sum of all the torques acting on the object is zero. As per Newton’s first law, if there is no torque acting on the system, the object will continue its state of rest or state of angular motion.

Formula:

$\mathrm{TLcos}\left(\mathrm{\theta }\right)-{\mathrm{F}}_{\mathrm{a}}\mathrm{y}=0$

With the pivot at the hinge, the net torque zero. It can be written as,

$\mathrm{TLcos}\left(\mathrm{\theta }\right)-{\mathrm{F}}_{\mathrm{a}}\mathrm{y}=0$

This leads to,

$\mathrm{T}=\left(\frac{{\mathrm{F}}_{\mathrm{a}}}{\mathrm{cos\theta }}\right)\left(\frac{\mathrm{y}}{\mathrm{L}}\right)$ …… (ii)

It can be interpreted${\mathrm{F}}_{\mathrm{a}}/\mathrm{cos\theta }$ as the slope on the tension graph. To get${\mathrm{F}}_{\mathrm{h}}$ use,

$\begin{array}{rcl}{\mathrm{F}}_{\mathrm{h}}& =& \mathrm{Tcos\theta }-{\mathrm{F}}_{\mathrm{a}}\\ & =& \left(-{\mathrm{F}}_{\mathrm{a}}\right)\left(\frac{\mathrm{y}}{\mathrm{L}}\right)-{\mathrm{F}}_{\mathrm{a}}\end{array}$

Substitute equation (ii), the result implies that the slope on the ${\mathrm{F}}_{\mathrm{h}}$ graph is equal to -f_a or f_a=300. Thus substituting this value we get,

$\mathrm{\theta }={60}^{\circ }$

The magnitude of the slope of F_h graph is equal to the magnitude of $\overrightarrow{{\mathrm{F}}_{\mathrm{a}}}$.

Consider point A and point B on F_h graph as shown in figure below.

The slope is calculated as,

$\begin{array}{rcl}\mathrm{Slope}& =& \frac{{\mathrm{B}}_{\mathrm{y}}-{\mathrm{A}}_{\mathrm{y}}}{{\mathrm{B}}_{\mathrm{x}}-{\mathrm{A}}_{\mathrm{x}}}\\ & =& \frac{0-120}{1-0.6}\\ & =& \frac{-120}{0.4}\\ & =& -300\end{array}$

Thus, the magnitude of$\overrightarrow{{\mathrm{F}}_{\mathrm{a}}}=300\mathrm{N}$