SOLUTION: A circular disc of cast iron is 7 ft. in diameter and 1 1/3 in. thick. Determine weight.
A plate 1 ft. square and 1 in. thick weighs 37.5 lbs.
Area of circle: pi * r^2.
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A plate 1 ft. square and 1 in. thick weighs 37.5 lbs.
Area of circle: pi * r^2.
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Question 1208118: A circular disc of cast iron is 7 ft. in diameter and 1 1/3 in. thick. Determine weight.
A plate 1 ft. square and 1 in. thick weighs 37.5 lbs.
Area of circle: pi * r^2.
Not sure how to solve. Found 3 solutions by ikleyn, math_tutor2020, greenestamps:Answer by ikleyn(52829) (Show Source):
You can put this solution on YOUR website! A circular disc of cast iron is 7 ft. in diameter and 1 1/3 in. thick. Determine weight.
A plate 1 ft. square and 1 in. thick weighs 37.5 lbs.
Area of circle: pi * r^2.
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Step by step
(1) Convert 7 ft to inches: 7 ft = 7*12 = 84 inches.
(2) Find the area of the circular disk in square inches using the formula.
Remember that the radius is half of the diameter.
(3) Find the volume of the circular disk in cubic inches by multiplying its area
by the thick, which is 4/3 inches.
(4) Find the density of the iron in by dividing 37.5 lbs by the area
of 1 ft by 1 ft plate, which is 12*12 = 144 square inches.
(5) To find weight, multiply the volume found in (3) by the density found in (4).
You can put this solution on YOUR website!
1 ft = 12 in
7 ft = 7*12 = 84 inches is the diameter, half that (42) is the radius
1 & 1/3 = 4/3
Volume of cylinder =
Volume of cylinder =
Volume of cylinder = cubic inches
"A plate 1 ft. square and 1 in. thick weighs 37.5 lbs"
means we have a 1 ft by 1 ft by 1 inch rectangular slab of iron
i.e. we have a 12 inch by 12 inch by 1 inch slab of iron
The volume of this would be length*width*height = 12*12*1 = 144 cubic inches.
144 cubic inches of iron weighs 37.5 pounds which will help set up a useful conversion factor.
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This is the exact weight of the circular disc in terms of pi.
Here are some select approximations of this weight when we try various approximations of pi
rounding precision
pi
weight
2
3.14
1923.25
3
3.142
1924.475
4
3.1416
1924.23
5
3.14159
1924.223875
6
3.141593
1924.2257125
7
3.1415927
1924.22552875
8
3.14159265
1924.225498125
9
3.141592654
1924.225500575
10
3.1415926536
1924.22550033
As we use more decimal digits of pi, the weight seems to approach roughly 1924.2255 pounds.
Start with the 37.5 pounds that the plate, 1 foot square and 1 inch thick, weighs.
The cross sectional area of the plate is 1 square foot.
The cross sectional area of the disc is .
The cross sectional area of the disc is greater than the cross sectional area of the plate by a factor of
The thickness of the plate is 1 inch; the thickness of the disc is 1 1/3 inches.
The disc is thicker than the plate by a factor of 1 1/3 = 4/3.
Multiply the weight of the plate by the factors by which the cross sectional area and thickness of the disc exceed those of the plate:
ANSWER: = 612.5pi = 1924 to the nearest whole number.