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
3
inches wide,
4
inches high, and
6
feet long can support a maximum of
28
tons. What is the maximum weight that could be supported by a beam that is
5
inches wide,
4
inches high, and
20
feet long?
translating your conditions:
W = k * w* h^2 / l
given: w = 3in, h = 4in, l = 6 ft or 72 in , W = 28 tons
28 = k(3)(16)/72
k = 42
W = 42 * w* h^2 / l
plugging in the second case:
W = 42(5)(16)/240 = 14 tons
To solve this problem, we can use the inverse variation formula:
w = k / (l × w × h^2)
where:
w is the maximum weight supported,
k is the constant of variation,
l is the length of the beam,
w is the width of the beam, and
h is the height of the beam.
Given that a beam that is 3 inches wide, 4 inches high, and 6 feet long can support a maximum of 28 tons, we can use these values to find the value of k.
Substituting these values into the equation, we get:
28 = k / (6 × 3 × 4^2)
Simplifying, we have:
28 = k / (6 × 3 × 16)
28 = k / (288)
k = 28 × 288
k = 8064
Now that we have the value of k, we can substitute the dimensions of the second beam (5 inches wide, 4 inches high, and 20 feet long) into the equation to calculate the maximum weight it can support:
w = 8064 / (20 × 5 × 4^2)
w = 8064 / (20 × 5 × 16)
w = 8064 / 1600
w = 5.04 tons
Therefore, the maximum weight that could be supported by a beam that is 5 inches wide, 4 inches high, and 20 feet long is approximately 5.04 tons.
To find the maximum weight that could be supported by a beam that is 5 inches wide, 4 inches high, and 20 feet long, we can use the inverse variation relationship and the given information about the first beam.
Let's denote the width, height, and length of the first beam as w1, h1, and l1, respectively. And let's denote the maximum weight it can support as m1.
According to the problem, we know that the maximum weight (m) varies inversely with the length (l) and jointly with the width (w) and the square of the height (h).
This can be written as:
m1 = k * (w1 / l1) * (h1^2)
where k is the constant of variation.
We can find the value of k by substituting the values of m1, w1, h1, and l1 from the given information of the first beam:
28 = k * (3 / 6) * (4^2)
28 = k * (1/2) * 16
28 = 8k
k = 28 / 8
k = 3.5
So, the equation for the maximum weight (m) becomes:
m = 3.5 * (w / l) * (h^2)
Now, we can calculate the maximum weight (m) that could be supported by the second beam with the given dimensions:
w = 5 inches, h = 4 inches, l = 20 feet
Converting 20 feet to inches, we get l = 20 * 12 = 240 inches.
Substituting these values into the equation:
m = 3.5 * (5 / 240) * (4^2)
m = 3.5 * (1 / 48) * 16
m = 0.1458 * 16
m ≈ 2.333 tons
Therefore, the maximum weight that could be supported by a beam that is 5 inches wide, 4 inches high, and 20 feet long is approximately 2.333 tons.