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When granular materials are allowed to fall freely, they form conical (cone-shaped) piles. The naturally occurring angle of slope, measured from the horizontal, at which the loose material comes to rest is called the angle of repose and varies for different materials. The angle of repose θ is related to the height h and base radius r of the conical pile by the equation θ = cot-1(r/h). A certain granular material forms a cone-shaped pile with a height of 19 feet and a base diameter of 30 feet. What is the height of a pile that has a base diameter of 104 feet?

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If I'm reading that right the cotangent^-1 can also be written as 1/tan^-1

the cone forms a triangle with height 19 feet and base 15 giving you a side angle side triangle with which to find the angle theta and if I'm reading the problem right the theta is the same for both piles....in which case the new pile will have angle (theta) base 52 (.5 *104) angle 90

I hope my answer has come to your help. Have a nice day ahead and may God bless you always!

Answer:

The given equation is

\(\theta = cot^{-1}(\frac{r}{h} )\)

Which is the definition of the angle of repose.

Remember that the angle of repose refers to the angle between the horizontal plane and the inclined side of the cone formed when we allow to fall freely granular materials.

We know by given

\(h=19\\r=\frac{d}{2}=\frac{30}{2}=15\)

Replacing these values, we have

\(\theta = cot^{-1}(\frac{15}{19} ) \approx 38.29 \°\)

Therefore, the angle of repose is 38.29° approximately.

Now, if the base diameter is 104 feet, the height would be

\(38.29 = cot^{-1}(\frac{52}{h}\\tan(38.29)=\frac{52}{h}\\ h=\frac{52}{0.79} \approx 65.8\)

Therefore, the height would be 65.8 feet, assuming that the angle of repose is the same.

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