How much does a wrecking ball weigh 6 ft diameter made of steel
I guess you are using English units so
density of steel = .291 pounds/ in^3 or 509 pound/ft^3
Radius = R = 6/2 = 3 ft
volume = (4/3) pi r^3 = 4 pi *9 = 36 pi = 113 ft^3
509 * 113 = weight in pounds (heavy)
To determine the weight of a wrecking ball with a 6 ft diameter made of steel, we need to know the density of steel and use basic mathematical calculations.
Step 1: Find the volume of the wrecking ball. The volume of a sphere can be calculated using the formula:
Volume = (4/3) * π * r^3
Where π is approximately 3.14159 and r is the radius of the wrecking ball, which is half of its diameter (6 ft).
Step 2: Convert the diameter to radius. The radius would be 6 ft divided by 2, which is 3 ft.
Step 3: Calculate the volume:
Volume = (4/3) * 3.14159 * (3 ft)^3
Volume = (4/3) * 3.14159 * 27 ft^3
Volume ≈ 113.097 ft^3 (rounded to three decimal places)
Step 4: Determine the density of steel. The density of steel can vary, but for this example, we'll assume a typical value of 490 pounds per cubic foot.
Step 5: Calculate the weight:
Weight = Volume * Density
Weight ≈ 113.097 ft^3 * 490 lb/ft^3
Weight ≈ 55,262.43 lb (rounded to two decimal places)
Therefore, a wrecking ball with a 6 ft diameter made of steel would weigh approximately 55,262 pounds.