Chapter 4.7, Problem 102E

(a)

To determine

The diagram that gives the visual representation of the problem.

Answer to Problem 102E

The values of OB is 4m and OA is 9.4m

Explanation of Solution

Given information:

To find the angle of repose for shelled corn and how tall is a pile of shelled com that has a base diameter of 12 meters.

Formula used:

The graph is plotted against the values.

Calculation:

A visual representation of the problem is

  

Conclusion:

The values of OB is 4m and OA is 9.4m

(b)

To determine

The angle of the response for shelled com.

Answer to Problem 102E

  ${23}^{\circ }$ be the angle of repose for shelled corn.

Explanation of Solution

Given information:

To find the angle of repose for shelled corn and how tall is a pile of shelled com that has a base diameter of 12 meters.

Formula used:

  $\tan \theta =\frac{OB}{OA}$

Calculation:

Let $\theta$ be the angle of repose for shelled corn. Here $OA=9.4\text{and }OB=4$ . Use the tangent formula in triangle AOB

  $\begin{array}{l}\tan \theta =\frac{OB}{OA}\\ =\frac{4}{9.4}\\ \theta =\arctan \left(\frac{4}{9.4}\right)\\ \theta ={23}^{\circ }\end{array}$

Thus, ${23}^{\circ }$ be the angle of repose for shelled corn.

Conclusion:

  ${23}^{\circ }$ be the angle of repose for shelled corn.

(c)

To determine

The height of the pile that is shelled com having the base diameter of 12 meters.

Answer to Problem 102E

The approx. 2.5 m tail is a pile of shelled com that has a base diameter of 12 meters.

Explanation of Solution

Given information:

To find the angle of repose for shelled corn and how tall is a pile of shelled com that has a base diameter of 12 meters.

Formula used:

  $\tan \theta =\frac{OB}{OA}$

Calculation:

When shelled corn has a base diameter of 12 meters so, $OA=\frac{12}{2}=6$ . Use the angle of part (b) and have

  $\begin{array}{l}\text{tanθ=}\frac{\text{OB}}{\text{OA}}\\ \text{tan}{\text{23}}^{\text{o}}\text{=}\frac{\text{OB}}{\text{6}}\\ \left(\frac{\text{4}}{\text{9}\text{.4}}\right)\left(\text{6}\right)\text{=OB}\\ \text{OB }\approx \text{2}\text{.5}\end{array}$

Thus, approx. 2.5 m tail is a pile of shelled com that has a base diameter of 12 meters.

Conclusion:

The approx. 2.5 m tail is a pile of shelled com that has a base diameter of 12 meters.

(c)

To determine

The tallness of the pile which has the base diameter of 60 feet.

Answer to Problem 102E

The 2.5 m tall is a pile of shelled corn that has a base diameter of 12 meters.

Explanation of Solution

Given information:

To find the angle of repose for shelled corn and how tall is a pile of shelled com that has a base diameter of 12 meters.

Formula used:

  $\tan \theta =\frac{OB}{OA}$

Calculation:

When shelled corn has a base diameter of 12 meters so, $OA=\frac{12}{2}=6$ . Use the angle of part (b) and have

  $\begin{array}{l}\tan \theta =\frac{OB}{OA}\\ \tan {23}^{\circ }=\frac{OB}{6}\\ \frac{4}{9.4}6=OB\\ OB=25\end{array}$

Thus, approx 2.5 m tall is a pile of shelled corn that has a base diameter of 12 meters.

Conclusion:

The 2.5 m tall is a pile of shelled corn that has a base diameter of 12 meters.