Chapter 6.2, Problem 55E

a)

To determine

To find: The relationship between $x\&\theta$ .

Answer to Problem 55E

The required equation is $x^2-3x\cos \theta =46.75$

Explanation of Solution

Given information: The engine has 7 inch long connecting rod that is fastened to a crank.

Calculation: By the help of the figure,

  $\begin{array}{l}{7}^2={1.5}^2+x^2-2\left(1.5\right)\left(x\right)\cos \theta \\ 49=2.25+x^2-3x\cos \theta \\ x^2-3x\cos \theta =46.75\end{array}$

Therefore, the required equation is $x^2-3x\cos \theta =46.75$ .

b)

To determine

To write: x as a function of $\theta$

Answer to Problem 55E

The required function is $x=\frac{1}{2}\left(3\cos \theta +\sqrt{9{\cos }^2\theta +187}\right)$

Explanation of Solution

Given information: The obtained equation of between $x\&\theta$ is $x^2-3x\cos \theta =46.75$

Calculation: Solve the $x^2-3x\cos \theta =46.75$ ,

  $\begin{array}{l}{x}^2-3x\cos \theta =46.75\\ x^2-3x\cos \theta +{\left(\frac{3\cos \theta }{2}\right)}^2=46.75+{\left(\frac{3\cos \theta }{2}\right)}^2\\ {\left[x-\frac{3\cos \theta }{2}\right]}^2=\frac{187}{4}+\frac{9{\cos }^2\theta }{4}\\ x-\frac{3\cos \theta }{2}=\sqrt{\frac{187}{4}+\frac{9{\cos }^2\theta }{4}}\end{array}$

According to the question, we have to consider only positive value of x,

  $x=\frac{1}{2}\left(3\cos \theta +\sqrt{9{\cos }^2\theta +187}\right)$

c)

To determine

To graph: The function $x=\frac{1}{2}\left(3\cos \theta +\sqrt{9{\cos }^2\theta +187}\right)$

Explanation of Solution

Given information: $x=\frac{1}{2}\left(3\cos \theta +\sqrt{9{\cos }^2\theta +187}\right)$

Graph: The graph of $x=\frac{1}{2}\left(3\cos \theta +\sqrt{9{\cos }^2\theta +187}\right)$

is as follows:

  

d)

To determine

To find: The total distance moved by the piston in one cycle.

Answer to Problem 55E

The total distance moved by the piston in one cycle is $6inches$ .

Explanation of Solution

Given information:

  

Calculation: By the help of the graph,

When $\theta =0\&\theta =2\pi$ ,

The distance $x=8.5$

And when $\theta =\pi$ ,

The distance will be $=2\left(8.5-5.5\right)$

  $\begin{array}{l}=2\times 3\\ =6inches\end{array}$