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