Chapter 11, Problem 13PS

a.

To determine

By using a graphing utility to graph the resulting function of1y $\theta$ .

Answer to Problem 13PS

  

Explanation of Solution

Given information:

In physics, the cross product can be used to measure torque, or the moment

  $M$ of a force $F$ about a point $P$ . If the point of application of the force is $Q$ then the moment of

  $F$ about $P$ is given by:

  $M=\overrightarrow{PQ}\times F$

A force of $60$ pounds acts on the pipe wrench shown in the figure.

Find the magnitude of the moment about $O$ . Use a graphing utility to graph the resulting function of $\theta$ .

Calculation:

Here, we will consider the following pipe wrench system as shown in the figure:

  

Now, the force acting at a point $A$ is of $60$ pounds. The moment about point $O$ is calculated as:

  $M=OA\times F$

So, the magnitude $\left|M\right|=\left|OA\right|\left|F\right|\sin \theta$ , coordinates of point $A$ are:

  $\begin{array}{l}x=18\times \cos 30°\\ =15.5in\\ y=18\times \sin 30°\\ =9in\\ z=0\end{array}$

Therefore, vector $\overrightarrow{OA}$ is:

  $\overrightarrow{OA}=\langle \left(15.5-0\right),\left(9-0\right),\left(0-0\right)\rangle$

  $=\langle 15.5,9,0\rangle$

The magnitude of this moment is:

  $M=\left|\langle 15.5,9,0\rangle \right|\left|\langle 0,-60,0\rangle \right|\sin \theta$

  $=\sqrt{{\left(15.5\right)}^2+{\left(9\right)}^2}\times 60\sin \theta$

  $=1075.4\sin \theta$

Now, by using $TI=83$ , the graph of moment about point $O$ can be plotted.

Now, press $\boxed{Y=}$ button and write the equation:

  

Now, press $\boxed{WINDOW}$ button and make such readings:

  

Now, press $\boxed{GRAPH}$ button and the graph is:

  

b.

To determine

To determine the magnitude of the moment when $\theta =45°$ .

Answer to Problem 13PS

  $\boxed{760.42Nm}$

Explanation of Solution

Given information:

In physics, the cross product can be used to measure torque, or the moment $M$ of a force $F$ about a point $P$ . If the point of application of the force is $Q$ then the moment of $F$ about $P$ is given by:

  $M=\overrightarrow{PQ}\times F$

A force of $60$ pounds acts on the pipe wrench shown in the figure.

Use the result of part (a) to determine the magnitude of the moment when $\theta =45°$ .

Calculation:

Here, we will put $\theta =45°$ in the equation $M=1075.4\sin \theta$ .

  $M=1075.4\sin 45°$

  $=760.42N-m$

Hence, at $\theta =45°$ the magnitude of the moment about $O$ is $\boxed{760.42Nm}$.

c.

To determine

To determine the angle $\theta$ when the magnitude of the moment is maximum.

Answer to Problem 13PS

  $\boxed{90°}$ , $\boxed{1075.4Nm}$

Explanation of Solution

Given information:

In physics, the cross product can be used to measure torque, or the moment $M$ of a force $F$ about a point $P$ . If the point of application of the force is $Q$ then the moment of $F$ about $P$ is given by:

  $M=\overrightarrow{PQ}\times F$

A force of $60$ pounds acts on the pipe wrench shown in the figure.

Use the result of part (a) to determine the angle $\theta$ when the magnitude of the moment is maximum. Is the answer what you expected? Why or why not?

Calculation:

Here, the slope of magnitude will be zero when magnitude is maximum.

Thus,

  $\begin{array}{l}\frac{d}{d\theta }\left(1075.4\sin {\theta }_{\max }\right)=0\\ \left(1075.4\cos {\theta }_{\max }\right)=0\\ {\theta }_{\max }={\cos }^{-1}0\\ {\theta }_{\max }=\boxed{90°}\end{array}$

Therefore, the maximum moment will be:

  $\begin{array}{l}{M}_{\max }=1075.4\sin {\theta }_{\max }\\ M_{\max }=1075.4\sin 90°\\ M_{\max }=1075.4Nm\end{array}$

Hence, the maximum moment is $\boxed{1075.4Nm}$.