Chapter 11, Problem 13PS

a.

To determine

To find: Magnitude of the moment about $O$ and graph the resulting function of $\theta$ .

Answer to Problem 13PS

The magnitude of the moment about $O$ is $\stackrel{\rightharpoonup }{OA}\times \text{F}=1080\sin \theta \text{}\text{inch pounds}$ .

By using graphing utility we get,

  

Explanation of Solution

Given: Force $=60\text{lbs}$ and acting downwards.

Angle from $O$ to $A$ is ${30}^{\circ }$ .

Distance from $O$ to $A$ is $18\text{in}$ .

  $\text{F}=-60\text{k}$

Distance from $O$ to $A$ is $18\text{in}$ .

The length of the pipe wrench is,

  $\stackrel{\rightharpoonup }{OA}=18\left(\cos {30}^{\circ }\text{j}+\sin {30}^{\circ }\text{k}\right)$

For a function $\theta$ the distance is,

  $\stackrel{\rightharpoonup }{OA}=18\left(\cos \theta \text{j}+\sin \theta \text{k}\right)$

By using the matrix we get,

  $\begin{array}{l}\stackrel{\rightharpoonup }{OA}\times \text{F=}\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& 18\sin \theta & 18\cos \theta \\ 0& 0& -60\end{array}\right|\\ \stackrel{\rightharpoonup }{OA}\times \text{F}=-60\left(18\sin \theta \right)\\ \stackrel{\rightharpoonup }{OA}\times \text{F}=-1080\sin \theta \\ \Vert \stackrel{\rightharpoonup }{OA}\times \text{F}\Vert =1080\sin \theta \text{ in-lbs}\end{array}$

Therefore the magnitude of the moment about $O$ is $\stackrel{\rightharpoonup }{OA}\times \text{F}=1080\sin \theta \text{}\text{inch pounds}$ .

b.

To determine

To find: The magnitude of the moment when $\theta ={45}^{\circ }$ .

Answer to Problem 13PS

The magnitude of the moment when $\theta ={45}^{\circ }$ is $\stackrel{\rightharpoonup }{OA}\times \text{F}=540\sqrt{2}\text{}\text{inch pounds}$ .

Explanation of Solution

Given: Force $=60\text{lbs}$ and acting downwards.

Angle from $O$ to $A$ is ${30}^{\circ }$ .

Distance from $O$ to $A$ is $18\text{in}$ .

From part a,

  $\stackrel{\rightharpoonup }{OA}\times \text{F}=1080\sin \theta$

At $\theta ={45}^{\circ }$ ,

  $\begin{array}{l}\stackrel{\rightharpoonup }{OA}\times \text{F}=1080\sin {45}^{\circ }\\ \stackrel{\rightharpoonup }{OA}\times \text{F}=540\sqrt{2}\end{array}$

Therefore the magnitude of the moment when $\theta ={45}^{\circ }$ is $\stackrel{\rightharpoonup }{OA}\times \text{F}=540\sqrt{2}\text{}\text{inch pounds}$ .

c.

To determine

To find: To find the angle $\theta$ when the magnitude of the moment is maximized.

Answer to Problem 13PS

The magnitude of the moment is maximum when $\theta ={90}^{\circ }$ .

Explanation of Solution

Given: Force $=60\text{lbs}$ and acting downwards.

Angle from $O$ to $A$ is ${30}^{\circ }$ .

Distance from $O$ to $A$ is $18\text{in}$ .

From ‘a’ we know that $\Vert \stackrel{\rightharpoonup }{OA}\times \text{F}\Vert =90\sin \theta$ ,

The maximum value of $\sin \theta =1$ , then the magnitude of the moment is maximum when $\theta ={\sin }^{-1}1={90}^{\circ }$ .

Therefore this is expected because $\theta ={90}^{\circ }$ .