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.)

Answer 1

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.