Chapter 11, Problem 14PS

a.

To determine

To find: The vector $\stackrel{\rightharpoonup }{AB}$ and the vector $\text{F}$ representing the force.

Answer to Problem 14PS

The vector of the force $\text{F=}-200\cos \theta \text{j}-200\sin \theta \text{k}\left(\text{lbs}\right)$ and the vector $\stackrel{\rightharpoonup }{AB}=-1.25\text{j}+\text{k}\left(\text{feet}\right)$

Explanation of Solution

Given: Force $=200\text{lbs}$ and acting downwards.

Locate the axes

  

Force is $\text{F=}-200\cos \theta \text{j}-200\sin \theta \text{k}\left(\text{lbs}\right)$ .

The vector $\stackrel{\rightharpoonup }{AB}=-1.25\text{j}+\text{k}\left(\text{feet}\right)$ .

Thus we can represent the force and the vector $\stackrel{\rightharpoonup }{AB}$ .

b.

To determine

To find: The magnitude of the moment about $A$ .

Answer to Problem 14PS

The magnitude of the moment about $A$ is $\Vert \stackrel{\rightharpoonup }{AB}\times \text{F}\Vert =250\sin \theta +200\cos ft-lb$ .

By using graphing utility,

  

Explanation of Solution

Given: Force $=200\text{lbs}$ and acting downwards.

From part a,

  $\text{F=}-200\cos \theta \text{j}-200\sin \theta \text{k}\left(\text{lbs}\right)$

  $\stackrel{\rightharpoonup }{AB}=-1.25\text{j}+\text{k}\left(\text{feet}\right)$

By using the matrix,

  $\begin{array}{l}\stackrel{\rightharpoonup }{AB}\times \text{F=}\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 0& -1.25& 1\\ 0& -200\sin \theta & -200\sin \theta \end{array}\right|\\ \stackrel{\rightharpoonup }{AB}\times \text{F=}\left(250\sin \theta +200\cos \theta \right)\text{i}\end{array}$

Therefore the magnitude of the moment $A$ is $\Vert \stackrel{\rightharpoonup }{AB}\times \text{F}\Vert =250\sin \theta +200\cos ft-lb$ .

c.

To determine

To find: To find the magnitude of moment when $\theta ={30}^{\circ }$ .

Answer to Problem 14PS

The magnitude of the moment when $\theta ={30}^{\circ }$ is $\Vert \stackrel{\rightharpoonup }{AB}\times \text{F}\Vert \approx 298.205ft-lb$ .

Explanation of Solution

Given: Force $=200\text{lbs}$ and acting downwards.

From ‘b’ we know that $\Vert \stackrel{\rightharpoonup }{AB}\times \text{F}\Vert =250\sin \theta +200\cos \theta ft-lb$ ,

When $\theta ={30}^{\circ }$ ,

  $\begin{array}{l}\Vert \stackrel{\rightharpoonup }{AB}\times \text{F}\Vert =250\sin {30}^{\circ }+200\cos {30}^{\circ }\\ \Vert \stackrel{\rightharpoonup }{AB}\times \text{F}\Vert =250\left(\frac{1}{2}\right)+200\left(\frac{\sqrt{3}}{2}\right)\\ \Vert \stackrel{\rightharpoonup }{AB}\times \text{F}\Vert \approx 298.205\end{array}$

Thus we can find the magnitude of moment when $\theta ={30}^{\circ }$ .

d.

To determine

To find: To determine the angle at the maximum magnitude of the moment.

Answer to Problem 14PS

The magnitude of the moment is maximized at $\theta \approx {53.34}^{\circ }$ .

Explanation of Solution

Given: Force $=200\text{lbs}$ and acting downwards.

From ‘b’ we know that $\Vert \stackrel{\rightharpoonup }{AB}\times \text{F}\Vert =250\sin \theta +200\cos \theta ft-lb$ ,

  

Thus we can find the magnitude of moment is maximized when $\theta \approx {53.34}^{\circ }$ .

e.

To determine

To find: To approximate the zero of the function.

Answer to Problem 14PS

The zero of the function is $\theta \approx {141.34}^{\circ }$ which is parallell to the vector $\stackrel{\rightharpoonup }{AB}$ .

Explanation of Solution

Given: Force $=200\text{lbs}$ and acting downwards.

From ‘b’ we know that $\Vert \stackrel{\rightharpoonup }{AB}\times \text{F}\Vert =250\sin \theta +200\cos \theta ft-lb$ ,

  

Thus we can approximate the zero of the function is $\theta \approx {141.34}^{\circ }$ which is parallell to the vector $\stackrel{\rightharpoonup }{AB}$ .