Chapter 6.3, Problem 116E

(a)

To determine

To Find: $\Vert F_1+F_2\Vert$ as a functionof $\theta$

Answer to Problem 116E

  $\Vert F_1+F_2\Vert =5\sqrt{5+4\cos \theta }$

Explanation of Solution

Given information:

Two forces

  $F_1=(10,0)\text{ and }{F}_2=5(\cos \theta ,\sin \theta )$

Concept used:

let $u=(u_1,u_2)$ and $v=(v_1,v_2)$ be vectors. Then the sum of $u\text{ and }v$ is the vector

  $u+v=(u_1+v_1,u_2+v_2)$

Calculation:

  $\Vert F_1+F_2\Vert$ is obtained as:

  $\begin{array}{l}{F}_1+F_2=(10+5\cos \theta ,0+5\sin \theta )\\ \Vert F_1+F_2\Vert =\sqrt{{\left(10+5\cos \theta \right)}^2+{\left(5\sin \theta \right)}^2}\\ \Vert F_1+F_2\Vert =\sqrt{100+25({\cos }^2\theta +{\sin }^2\theta )+100\cos \theta }\end{array}$

Since, ${\cos }^2\theta +{\sin }^2\theta =1$

  $\begin{array}{l}\Vert F_1+F_2\Vert =\sqrt{100+25+100\cos \theta }\\ \Vert F_1+F_2\Vert =\sqrt{125+100\cos \theta }\\ \Vert F_1+F_2\Vert =5\sqrt{5+4\cos \theta }\end{array}$

(b)

To determine

To graph: $5\sqrt{5+4\cos \theta }$ for $0\le \theta \le 2\pi$

Explanation of Solution

Given information:

Use a graphing utility to graph of the function $5\sqrt{5+4\cos \theta }$ for $0\le \theta \le 2\pi$

Graph:

Use graphing calculator to graph the function and the displayed graph is shown below:

  

(c)

To determine

The range of the function $5\sqrt{5+4\cos \theta }$

Answer to Problem 116E

the range of the function is $\left[5,15\right]$

Explanation of Solution

Given information:

The graph of the function $5\sqrt{5+4\cos \theta }$ for $0\le \theta \le 2\pi$

  

Calculation:

The maximum value is at $15$ and it occurs at $\theta =0$ and the minimum value of the function is $3$ and it occurs at $\theta =\pi$

So, the range of the function is $\left[5,15\right]$

(d)

To determine

To Explain:Why the magnitude of the resultant is never 0

Explanation of Solution

Given information:

The function $5\sqrt{5+4\cos \theta }$ for $0\le \theta \le 2\pi$

Concept Used: Range of $\cos \theta =\left[-1,1\right]$

Calculation:

The function is $5\sqrt{5+4\cos \theta }$ for $0\le \theta \le 2\pi$

  $\begin{array}{l}5+4\cos \theta =0\\ \cos \theta =\frac{-5}{4}\end{array}$

Range of $\cos \theta =\left[-1,1\right]$ and

  $\cos \theta =\frac{-5}{4}$ is beyond the range of a $\cos$ function, therefore the resultant cannot be zero.