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, l. A wooden beam 3in. wide, 6in. deep, and 11ft long holds up 1213lb. What load would a beam 6in. wide, 3in. deep and 12ft long of the same material support? (Round off your answer to the nearest pound.)

L' = 555.95 lb

Explanation:

Analyzing the given conditions in the question, we get

The safe load, L is directly proportional to width (w) and square of depth (d²)

also,

L is inversely proportional length (l) i. e L = k/l

combining the above conditions, we get an equation as:

L = k (wd²/l)

now, for the first case we have been given

w = 3 in

d = 6 in

l = 11 ft

L = 1213 lbs

thus,

1213 lb = k ((3 * 6²) / 11)

or

k = 123.54 lbs / (ft. in³)

Now,

Using the calculated value of k to calculate the value of L in the second case

in the second case, we have

w = 6 in

d = 3 in

l = 12 ft

Final Safe load L' = 123.54 * (6 * 3²/12)

or

L' = 555.95 lb