The safe load, L. of a wooden beam supported at both ends varies jointly as the width, w, the square of the depth, d, and inversely as the length, I. A wooden beam 7 in. wide. 4 in. deep, and 14 ft long holds up 9036 Ib. What load would a beam 6 in. wide, 8 in. deep and 11 ft long of the same material support? (Round off your answer to the nearest pound.)
We can set up a proportion to solve this problem.
Let L1 be the safe load of the first beam (7 in. wide, 4 in. deep, and 14 ft long) and L2 be the safe load of the second beam (6 in. wide, 8 in. deep, and 11 ft long).
According to the problem, L1 varies jointly as w, d^2, and inversely as I. So we can write the equation:
L1 = k * w * d^2 / I,
where k is the constant of variation. We can solve for k by plugging in the values given: L1 = 9036 lb, w = 7 in, d = 4 in, and I = 14 ft.
9036 = k * 7 * 4^2 / 14,
k = 9036 * 14 / (7 * 16),
k = 9036 * 2 / 16,
k = 1129.5.
Now we can use this value of k to find L2. Plugging in the values for the second beam: L2 = k * 6 * 8^2 / 11.
L2 = 1129.5 * 6 * 8^2 / 11,
L2 = 1129.5 * 6 * 64 / 11,
L2 = 54,629.45 / 11,
L2 ≈ 4975.4 lb.
Therefore, the load that the second beam can support is approximately 4975 lb when rounded to the nearest pound.