Chapter 5, Problem 9PS

a.

To determine

Simplify the model.

Answer to Problem 9PS

  $\boxed{F=2.8858W\cos \theta }$

Explanation of Solution

Given information:

The force $F$ (in pounds) on a person’s back when he or she bends over at an angle $\theta$ from an upright position is modelled by

  $F=\frac{0.6W\sin \left(\theta +90°\right)}{\sin 12°}$

where $W$ represents the person’s weight (in pounds).

Calculation:

Consider that force $F$ (in pounds)on a person’s back when he is bent over an angle $\theta$ is modelled by,

  $F=\frac{0.6W\sin \left(\theta +90°\right)}{\sin 12°}$

Here $W$ is person’s weight.

Use the following formula,

  $\sin \left(A+B\right)=\sin A\cos B+\cos A\sin B$

And the value, $\sin 12°=0.2079$

Then,

  $\sin \left(\theta +90°\right)=\sin \theta \cos 90°+\cos \theta \sin 90°$

  $\begin{array}{l}=\sin \theta .(0)+\cos \theta .(1)\\ =\cos \theta \end{array}$

Therefore force, $F$ is,

  $\begin{array}{l}F=\frac{0.6W\cos \theta }{0.2079}\\ =2.8858W\cos \theta \end{array}$

Hence, simplified form is,

  $\boxed{F=2.8858W\cos \theta }$

b.

To determine

Use a graphing utility to graph the model.

Answer to Problem 9PS

  

Explanation of Solution

Given information:

The force $F$ (in pounds) on a person’s back when he or she bends over at an angle $\theta$ from an upright position is modelled by

  $F=\frac{0.6W\sin \left(\theta +90°\right)}{\sin 12°}$

where $W$ represents the person’s weight (in pounds).

Calculation:

Consider that force $F$ (in pounds)on a person’s back when he is bent over an angle $\theta$ is modelled by,

  $F=\frac{0.6W\sin \left(\theta +90°\right)}{\sin 12°}$

Here $W$ is person’s weight.

Consider the value of weight, $W=185$

It gives,

  $F=533.88\cos \theta$

Use the T1-83 calculator to graph the model.

Enter the key $"Y"$ and enter the equation as shown below,

  

Set $\theta$ as $0<\theta <90$ by clicking on the key “WINDOW”

  

Click on the key “GRAPH”

Graph of the model will appear as shown below,

  

c.

To determine

At what angle is the force maximized? At what angle is the force minimized?

Answer to Problem 9PS

Maximum at $\boxed{\theta =0°}$ and minimum at $\boxed{\theta =90°}$

Explanation of Solution

Given information:

The force $F$ (in pounds) on a person’s back when he or she bends over at an angle $\theta$ from an upright position is modelled by

  $F=\frac{0.6W\sin \left(\theta +90°\right)}{\sin 12°}$

where $W$ represents the person’s weight (in pounds).

Calculation:

Consider that force $F$ (in pounds)on a person’s back when he is bent over an angle $\theta$ is modelled by,

  $F=\frac{0.6W\sin \left(\theta +90°\right)}{\sin 12°}$

Here $W$ is person’s weight.

Observe the graph.

It is clear that the force is maximum at angle $0$ degree and minimum at angle $90$ degree.

Force is maximum at $\boxed{\theta =0°}$ and minimum at $\boxed{\theta =90°}$