SOLUTION: A steel gear is to be lightened by drilling holes through the gear. The gear is 3.27 in. thick. Find the diameter of the holes if each is to remove 12 oz. Use a density 485 lb/ft3

Question 906184: A steel gear is to be lightened by drilling holes through the gear. The gear is 3.27 in. thick. Find the diameter of the holes if each is to remove 12 oz. Use a density 485 lb/ft3 for iron.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
density is 485 pounds per cubic foot.
a pound is equal to 16 ounces
a foot is equal to 12 ounces.
since a cubic foot is 1*1*1, then the same volume in inches is 12*12*12 = 1728 cubic inches.
bottom line is 1 cubic foot is equal to 1728 cubic inches.
since a pound is equal to 16 ounces, then 485 pounds is equal to 485 * 16 = 7760 ounces.

what you get is that 485 pounds per cubic foot is equivalent to 7760 ounces per 1728 cubic inches which can be simplified to 4.49... ounces per cubic inch.

now the hole you drill will be 3.27 inches in height because that's the thickness of the gear.

you want to find the diameter of the hole required to remove 12 ounces of iron that the gear is presumably made of.

12 ounces at 4.49... ounces per cubic inch will be equal to 12 * 4.49... = 53.88... cubic inches.

you want to remove 53.88... cubic inches of iron.

the formula for the volume of a cylinder is v = pi*r^2*h
we know that v = 53.88... cubic inches and we know that h = 3.27 inches, so we can solve for r to get:

53.88... = pi * r^2 * 3.27, and when we solve for r we get r = sqrt(53.88.../(3.27*pi) which makes r = 2.29... inches.

since the radius is 2.29... inches, then the diameter is twice that, so the diameter is 5.58... inches.

i believe that's your answer.

the diameter of the hole needs to be 4.58 inches rounded to 2 decimal places.

all calculations used the exact numbers provided by the calculator with rounding only done at the end.

the exact answer, to the accuracy of the calculator, is that the diameter needs to be 4.580688901 inches.

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