Suppose that the maximum weight that a certain type of rectangular beam can support varies inversely as its length and jointly as its width and the square of its height. Suppose also that a beam 4 inches wide, 3 inches high, and 18 feet long can support a maximum of 3 tons. What is the maximum weight that could be supported by a beam that is 7 inches wide, 4 inches high, and 6 feet long?
Uing f for weight (force), we have
f = kwh^2/l
So,
fl/(wh^2) is constant
You need f such that
6f/(7*4^2) = (3*18)/(4*3^2)
f = 28 tons
The maximum weight that a rectangular beam can support varies jointly as its width and the square of its height and inversely as its length. If a beam 1/3 foot wide, 1/3 foot high, and 16 feet long can support 30 tons, find how much a similar beam can support if the beam is 3/4 foot wide, 1/2 foot high, and 20 feet long
Given information:
Width (w1) = 4 inches
Height (h1) = 3 inches
Length (l1) = 18 feet
Weight (W1) = 3 tons
Using the inverse variation formula:
W1 = k / (w1 * h1^2 * l1)
where k is the constant of variation.
Now, we need to find the value of k.
Rearranging the formula, we get:
k = W1 * w1 * h1^2 * l1
Plugging in the given values:
k = 3 tons * 4 inches * (3 inches)^2 * 18 feet
To proceed with the calculation, we need to convert the units to a consistent system. Converting inches to feet, we have:
k = 3 tons * 4/12 feet * (3/12 feet)^2 * 18 feet
Simplifying the expression:
k = 3 tons * 1/3 feet * (1/4 feet)^2 * 18 feet
k = 1 ton * 1/3 * 1/16 * 18 feet^2
k = 1/3 * 1/16 * 18 tons * feet^2
k = 3/16 tons * feet^2
Now, we can find the maximum weight (W2) for a new set of dimensions:
Width (w2) = 7 inches
Height (h2) = 4 inches
Length (l2) = 6 feet
Using the inverse variation formula:
W2 = k / (w2 * h2^2 * l2)
Plugging in the known value for k, and the given dimensions:
W2 = (3/16 tons * feet^2) / (7 inches * (4 inches)^2 * 6 feet)
Converting inches to feet:
W2 = (3/16 tons * feet^2) / (7/12 feet * (4/12 feet)^2 * 6 feet)
Simplifying the expression:
W2 = (3/16 tons * feet^2) / (7/12 * 1/16 * 6 tons * feet^2)
W2 = (3/16 tons * feet^2) / (7/192 tons * feet^2)
W2 = (3/16 tons * 192 tons * feet^2) / (7 * feet^2)
W2 = (3 * 192 tons * feet^2) / (16 * 7 * feet^2)
W2 = (576 tons * feet^2) / (112 feet^2)
W2 = 576/112 tons
W2 ≈ 5.14 tons
Therefore, the maximum weight that could be supported by the beam with dimensions 7 inches wide, 4 inches high, and 6 feet long is approximately 5.14 tons.
To find the maximum weight that could be supported by a beam with different dimensions, we need to use the inverse variation formula.
Let's denote the maximum weight as W, the length as L, the width as W, the height as H, and the constant of variation as k.
According to the problem statement, we know:
- The maximum weight W varies inversely with the length L. This means that W ∝ 1/L.
- The maximum weight W varies jointly with the width W and the square of the height H. This means that W ∝ W * H^2.
Combining the two variations, we have the equation:
W ∝ (1/L) * (W * H^2)
To find the constant of variation k, we can substitute the given dimensions into the equation and solve for k.
For the first beam:
W = 3 tons
L = 18 feet (since 1 ton is equal to 2000 pounds and 1 ton = 2000 * 224 = 448,000 pounds, we have 3 tons = 3 * 448,000 = 1,344,000 pounds)
W = 4 inches
H = 3 inches
Substituting the values, we have:
1,344,000 = k * (1/18) * (4 * 3^2)
1,344,000 = k * (1/18) * (4 * 9)
1,344,000 = k * (1/18) * 36
k = 1,344,000 / (1/18 * 36)
k = 1,344,000 / (1/18 * 36)
k ≈ 1,344,000 / 6
k ≈ 224,000
Now that we have the constant of variation, we can find the maximum weight for the second beam.
For the second beam:
L = 6 feet
W = 7 inches
H = 4 inches
Substituting the values and using the constant of variation k, we have:
W = k * (1/L) * (W * H^2)
W = 224,000 * (1/6) * (7 * 4^2)
W = 224,000 * (1/6) * (7 * 16)
W = 224,000 * (1/6) * 112
W ≈ 448,000 pounds
Therefore, the maximum weight that could be supported by the beam with dimensions 7 inches wide, 4 inches high, and 6 feet long is approximately 448,000 pounds.