Question

The safe load, L, of a wooden beam supported at both ends varies jointly as the width, w, and the square of the depth, d, and inversely as the length, l. A wooden beam 5in.
5
⁢
in.
wide, 4in.
4
⁢
in.
deep, and 2ft
2
⁢
ft
long holds up 34387lb
34387
⁢
lb
. What load would a beam 8in.
8
⁢
in.
wide, 2in.
2
⁢
in.
deep, and 15ft
15
⁢
ft
. long, of the same material, support? Round your answer to the nearest integer if necessary

Answer 1

Answer:

joint variation means L (safe load) is directly proportional to both w and d2

inverse proportionality means L = k/l

So the equation is:

L = k(wd2/l)

You are given:

w = 6 in

d = 5 in

l = 12 ft

L = 7666 lbs

7666 = k(6*52/12)

Solve for the constant, k.

7666 = k (6*25/12)

7666 = k (150/12)

7666 = k (12.5)

613.28 lbs/(ft.in3) = k

Use this k-value to solve for L in the last part of the question.

What safe load, L would a beam 3 in. wide, 7 in. deep and 15 ft long of the same material support? (Round off your answer to the nearest pound.)

w = 3 in

d =7 in

l = 15 ft

k = 613.28 lbs/(ft.in3)

Final Safe load = 6010.144 lbs *** Edited for clarity and to fix a multiplication erro