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Expert-verified Found in: Page 348 ### Fundamentals Of Physics

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 $\stackrel{\mathbf{\to }}{{\mathbf{F}}_{\mathbf{a}}}$ 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 ${\mathbf{\theta }}$ with the horizontal. Figure 12-49a gives the tension T in the cable as a function of the position of the applied force given as a fraction y/L of the beam length. The scale of the T axis is set by ${{\mathbf{T}}}_{{\mathbf{s}}}{\mathbf{=}}{\mathbf{600}}{\mathbf{}}{\mathbf{N}}$. Figure gives the magnitude Fh of the horizontal force on the beam from the hinge, also as a function of y/L. Evaluate (a) angle ${\mathbf{\theta }}$ and (b) the magnitude of $\stackrel{\mathbf{\to }}{{\mathbf{F}}_{\mathbf{a}}}$. 1. Angle $\mathrm{\theta }\mathrm{is}{60}^{\circ }$
2. Magnitude of $\stackrel{\to }{{\mathrm{F}}_{\mathrm{a}}}$ is 300 N
See the step by step solution

## Step 1: Determine the given quantities

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

## Step 2: Determine the concept and the formulas:

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:

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

## Step 3: (a) To calculate angle

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

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

$\mathrm{T}=\left(\frac{{\mathrm{F}}_{\mathrm{a}}}{\mathrm{cos\theta }}\right)\left(\frac{\mathrm{y}}{\mathrm{L}}\right)$ …… (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 -fa or fa =300. Thus substituting this value we get,

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

## Step 4: (b) Calculate the magnitude of Fa→

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

Consider point A and point B on Fh 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 $\stackrel{\to }{{\mathrm{F}}_{\mathrm{a}}}=300\mathrm{N}$ ### Want to see more solutions like these? 