Chapter 6.3, Problem 103E

(a)

To determine

To find: The direction and magnitude of the resultant of the forces when $\theta ={30}^{\circ }.$

Answer to Problem 103E

Magnitude $=357.85\text{ N}$

Direction $={12}^{\circ }$

Explanation of Solution

Given: Forces with magnitudes of 150 newtons and 220 newtons act on a book, as shown in the figure.

  

First make the vertical and horizontal component of both forces.

For force 1: $F_1=150\cos \theta i+150\sin \theta j$

For force 2: $F_2=220i+0j$

The resultant of both forces, $F=\left(150\cos \theta +220\right)i+150\sin \theta j$

For $\theta ={30}^{\circ },$ magnitude of F

  $\begin{array}{l}\left|F\right|=\sqrt{{\left(150\cos {30}^{\circ }+220\right)}^2+{\left(150\sin {30}^{\circ }\right)}^2}\\ \left|F\right|=\sqrt{{\left(75\sqrt{3}+220\right)}^2+{\left(75\right)}^2}\\ \left|F\right|=357.85\text{ N}\end{array}$

Direction of F, $\alpha ={\tan }^{-1}\frac{150\sin {30}^{\circ }}{220+150\cos {30}^{\circ }}={12}^{\circ }$

Hence, the magnitude of resultant forces is 357.85 N and direction ${12}^{\circ }$

(b)

To determine

To find: The magnitude and direction in form of $\theta$ .

Answer to Problem 103E

  $\left|M\right|=10\sqrt{709+660\cos \theta }$

  $\alpha ={\tan }^{-1}\left(\frac{15\sin \theta }{22+15\cos \theta }\right)$

Explanation of Solution

Given: Forces with magnitudes of 150 newtons and 220 newtons act on a book, as shown in the figure.

  

First make the vertical and horizontal component of both forces.

For force 1: $F_1=150\cos \theta i+150\sin \theta j$

For force 2: $F_2=220i+0j$

The resultant of both forces, $M=\left(150\cos \theta +220\right)i+150\sin \theta j$

Magnitude of forces, $\left|M\right|=\sqrt{{\left(220+150\cos \theta \right)}^2+{\left(150\sin \theta \right)}^2}$

  $\begin{array}{l}\left|M\right|=\sqrt{{220}^2+{150}^2{\cos }^2\theta +{150}^2{\sin }^2\theta +2\cdot 220\cdot 150\cos \theta }\\ \left|M\right|=10\sqrt{709+660\cos \theta }\end{array}$

Let direction of resultant force is $\alpha$

  $\begin{array}{l}\alpha ={\tan }^{-1}\left(\frac{150\sin \theta }{220+150\cos \theta }\right)\\ \alpha ={\tan }^{-1}\left(\frac{15\sin \theta }{22+15\cos \theta }\right)\end{array}$

(c)

To determine

To complete: The table for different value of $\theta$ .

Answer to Problem 103E

  $\begin{array}{cccccccc}\theta & 0^{\circ }& {30}^{\circ }& {60}^{\circ }& {90}^{\circ }& {120}^{\circ }& {150}^{\circ }& {180}^{\circ }\\ M& 370& 358& 322& 266& 195& 117& 70\\ \alpha & 0^{\circ }& {12}^{\circ }& {24}^{\circ }& {34}^{\circ }& {42}^{\circ }& {40}^{\circ }& 0^{\circ }\end{array}$

Explanation of Solution

Given: Forces with magnitudes of 150 newtons and 220 newtons act on a book, as shown in the figure.

The magnitude of resultant forces, $\left|M\right|=10\sqrt{709+660\cos \theta }$

The direction of the resultant forces, $\alpha ={\tan }^{-1}\left(\frac{15\sin \theta }{22+15\cos \theta }\right)$

Using graphing utility to complete the table.

  $\begin{array}{cccccccc}\theta & 0^{\circ }& {30}^{\circ }& {60}^{\circ }& {90}^{\circ }& {120}^{\circ }& {150}^{\circ }& {180}^{\circ }\\ M& 370& 358& 322& 266& 195& 117& 70\\ \alpha & 0^{\circ }& {12}^{\circ }& {24}^{\circ }& {34}^{\circ }& {42}^{\circ }& {40}^{\circ }& 0^{\circ }\end{array}$

(d)

To determine

To graph: The function of magnitude and direction of resultant forces.

Answer to Problem 103E

  

Explanation of Solution

Given: Forces with magnitudes of 150 newtons and 220 newtons act on a book, as shown in the figure.

Using graphing utility to graph the function.

The magnitude of resultant forces, $\left|M\right|=10\sqrt{709+660\cos \theta }$

The direction of the resultant forces, $\alpha ={\tan }^{-1}\left(\frac{15\sin \theta }{22+15\cos \theta }\right)$

Where , y -axis represents magnitude and direction and x -axis represents angle $\theta$

  

(e)

To determine

To graph: The function of magnitude and direction of resultant forces.

Explanation of Solution

Given: Forces with magnitudes of 150 newtons and 220 newtons act on a book, as shown in the figure.

Using graphing utility to graph the function.

The magnitude of resultant forces, $\left|M\right|=10\sqrt{709+660\cos \theta }$

The direction of the resultant forces, $\alpha ={\tan }^{-1}\left(\frac{15\sin \theta }{22+15\cos \theta }\right)$

Where , y -axis represents magnitude and direction and x -axis represents angle $\theta$

  

The graph of magnitude of resultant of two forces is decreases as $\theta$ increasing because the magnitude depends on $\sqrt{\cos \theta }$ which is decreasing over $0\le \theta \le {180}^{\circ }$ .

The graph of direction of resultant of two forces is increase as $\theta$ increasing because the function of direction depends on $\theta$ which is increasingover $0\le \theta \le {180}^{\circ }$ .