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Found in: Page 347

Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

In Fig12-46, a ${\mathbf{50}}{\mathbf{.0}}{\mathbf{\text{\hspace{0.17em}kg}}}$ uniform square sign, of edge length${\mathbit{L}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{2}}{\mathbf{.00}}{\mathbf{\text{\hspace{0.17em}m}}}$ , is hung from a horizontal rod of length ${{\mathbit{d}}}_{{\mathbf{h}}}{\mathbf{=}}{\mathbf{}}{\mathbf{3}}{\mathbf{.00}}{\mathbf{\text{\hspace{0.17em}m}}}$ and negligible mass. A cable is attached to the end of the rod and to a point on the wall at distance ${{\mathbit{d}}}_{{\mathbf{v}}}{\mathbf{=}}{\mathbf{}}{\mathbf{4}}{\mathbf{.00}}{\mathbf{}}{\mathbf{\text{\hspace{0.17em}m}}}$ above the point where the rod is hinged to the wall.(a) What is the tension in the cable? What are the (b) magnitude and) of the horizontal component of the force on the rod from the wall, and the (c) direction (left or right) of the horizontal component of the force on the rod from the wall, and the (d) magnitude of the vertical component of this force? And (e) direction (up or down) of the vertical component of this force?

a) Tension in the cable is,$T=408\text{\hspace{0.17em}N}$ .

b) The magnitude of the horizontal component of the force on the rod from the wallis,${F}_{x}=245\text{\hspace{0.17em}N}$ .

c) The direction of the horizontal component of the force on the rod from the wallis towards the right.

d) The magnitude of the vertical component of the force on the rod from the wall is,${F}_{y}=163\text{\hspace{0.17em}N}$

.

e) The vertical component of the force on the rod from the wall is in the upward direction.

See the step by step solution

Step 1: Understanding the given information

i) Mass of the sign,$\text{}m=50.0\text{\hspace{0.17em}kg}$ .

ii) Length of the rod,${d}_{h}=3.00\text{\hspace{0.17em}m}$ .

iii) Distance from hinge to point where cable attached to the wall, ${d}_{v}=4.00\text{\hspace{0.17em}m}$

Step 2: Concept and formula used in the given question

Using the condition for equilibrium,you can write the torque equation. From this,you can find the tension in the cable. Then using the conditions for the equilibrium for horizontal and vertical forces,you can find the magnitude and direction of the horizontal and vertical components of the force acting on the rod from the wall.

Step 3: (a) Calculation for the tension in the cable

You can find the angle between the cable and the rod at which the sign is attached:

$\mathrm{tan}\theta =\frac{{d}_{v}}{{d}_{h}}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}\mathrm{tan}\theta =\frac{4.00}{3.00}\\ =1.33\\ \theta ={\mathrm{tan}}^{-1}\left(1.33\right)\\ ={53}^{\circ }\end{array}$

We can find the tension in the cable by using the condition for equilibrium as:

$\begin{array}{c}{\tau }_{net}=0\\ T\mathrm{sin}\theta \left({d}_{h}\right)-\left(\frac{mg}{2}\right)\left({d}_{h}-L\right)-\left(\frac{mg}{2}\right){d}_{h}=0\\ T\mathrm{sin}\theta \left(d\right)=\left(\frac{mg}{2}\right)\left({d}_{h}-L\right)+\left(\frac{mg}{2}\right){d}_{h}\\ T=\frac{\left(\frac{mg}{2}\right)\left({d}_{h}-L\right)+\left(\frac{mg}{2}\right){d}_{h}}{\mathrm{sin}\theta \left(d\right)}\end{array}$

Substitute the values in the above expression, and we get,

localid="1661236210391" $\begin{array}{c}T=\frac{\left(\frac{\left(50.0\right)\left(9.8\right)}{2}\right)\left(3.00-2.00\right)+\left(\frac{\left(50.0\right)\left(9.8\right)}{2}\right)\left(3.00\right)}{\left(\mathrm{sin}{53}^{\circ }\right)\left(3.00\right)}\\ =408\text{\hspace{0.17em}N}\end{array}$

Thus, the tension in the cable is, $T=408\text{\hspace{0.17em}N}$.

Step 4: (b) Calculation for the magnitude of the horizontal component of the force on the rod from the wall

At equilibrium,

${F}_{xnet}=0$

Substitute the terms in the above expression, and we get,

${F}_{x}-T\mathrm{cos}\theta =0$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{F}_{x}-T\mathrm{cos}{53}^{\circ }=0\\ {F}_{x}=T\mathrm{cos}{53}^{\circ }\\ =408×\mathrm{cos}{53}^{\circ }\\ =245\text{\hspace{0.17em}N}\end{array}$

Thus, the magnitude of the horizontal component of the force on the rod from the wall is, ${F}_{x}=245\text{\hspace{0.17em}N}$ .

Step 5: (c) Calculation for the direction (left or right) of the horizontal component of the force on the rod from the wall

From the figure, we can see that direction of the horizontal component of the force on the rod from the wall is towards the right.

Thus, the direction of the horizontal component of the force on the rod from the wall is towards the right.

Step 6: (d) Calculation for the magnitude of the vertical component of this force

At equilibrium,

${F}_{ynet}=0$

Substitute the terms in the above expression, and we get,

role="math" localid="1661236093581" $\begin{array}{c}{F}_{y}+T\mathrm{sin}\theta -mg=0\\ {F}_{y}=mg-T\mathrm{sin}\theta \end{array}$

Substitute the values in the above expression, and we get,

Thus, the magnitude of the vertical component of the force on the rod from the wall is,${F}_{y}=163\text{\hspace{0.17em}N}$ .

Step 7: (e) Calculation for the direction (up or down) of the vertical component of this force

The direction of the vertical component of the force on the rod from the wall is upward. Thus, the vertical component of the force on the rod from the wall is in the upward direction.

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