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 6 in. wide. 8 in. deep, and 16 ft long holds up 8420 Ib. What load would a beam 7 in. wide, 9 in. deep and 12 ft long of the same material support? (Round off your answer to the nearest pound.)
We can write the equation for the safe load, L, of the wooden beam as:
L = k * w * d^2 / I
where k is the constant of variation.
We can use the given information to find the value of k.
For the first beam with width, w = 6 in., depth, d = 8 in., and length, I = 16 ft, the safe load, L = 8420 lb.
8420 = k * 6 * 8^2 / 16
Solving for k, we get k ≈ 131.56.
Now, we can plug this value of k into the equation to find the safe load for the second beam with width, w = 7 in., depth, d = 9 in., and length, I = 12 ft.
L = 131.56 * 7 * 9^2 / 12
L ≈ 131.56 * 7 * 81 / 12
L ≈ 6493.65 lb
Therefore, the load that a beam 7 in. wide, 9 in. deep, and 12 ft long of the same material can support is approximately 6494 lb.