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33P

Expert-verifiedFound in: Page 348

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Figure 12-49a shows a vertical uniform beam of length L that is hinged at its lower end. A horizontal force $\overrightarrow{{\mathbf{F}}_{\mathbf{a}}}$ is applied to the beam at distance y from the lower end. The beam remains vertical because of a cable attached at the upper end, at angle ${\mathbf{\theta}}$**

- Angle $\mathrm{\theta}\mathrm{is}{60}^{\circ}$
- Magnitude of $\overrightarrow{{\mathrm{F}}_{\mathrm{a}}}$ is 300 N

The scale of the *T *axis 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 $\overrightarrow{{\mathrm{F}}_{\mathrm{a}}}=300\mathrm{N}$

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