Answer:
Load (L) = 156 lbs
Explanation:
Joint variation combines direct and inverse variation, so it's helpful to know the general forms of the direct and inverse variaition equations:
The general direct variation equation:
The general equation for direct variation is given by:
y = kx, where
- y varies directly as x,
- and k is the constant of proportionality.
The general inverse variation equation:
The general equation for inverse variation is given by:
y = k/x, where
- y varies inversely as x,
- and k is the constant of proportionality.
Creating the joint variation equation:
Since 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, we can represent this with the following equation:
L = (kwd^2) / l, where
- k is the constant of proportionality:
Determining the constant of proportionality, k:
Before we can determine what load would a beam 4 in. wide, 2 in. deep, and 13 ft. long of the same material would support, we first need to know the constant of proportionality, k.
We can find it by plugging in 3337 for L, 5 for w, 10 for d, and 19 for l, and solving for k:
(3337 = (k * 5 * 10^2) / 19) * 19
63403 = 5k * 100
(63403 = 500k) / 500
126.806 = k
Thus, the constant of proportionality is 126.806
Note that I didn't round k since it's better not to round until we get to the very end of the problem to get a more exact answer.
Determining the load supported by a beam 4 in. wide, 2 in. deep, and 13 ft. long:
Now we can find the load, L, supported by the beam that is 4 in. wide, 2 in. deep, and 13 ft. long by plugging in 126.806 for k, 4 for w, 2 for d, and 13 for l:
L = (126.806 * 4 * 2^2) / 13
L = (507.224 * 4) / 13
L = 156.0689231
L = 156
Thus, a load that is 156 lbs would be supported by a beam 4 in. wide, 2 in. deep, and 13 ft. long of the same material.