Answer :
Using the given joint variation equation, the safe load for a beam 18 inches in length made from the same material is determined to be 352 pounds, hence none of the provided options are correct.
The load p varies jointly with the product of the width w and the square of the depth d, and inversely with the length l. This relationship can be represented by the equation
p = k * (w * d^2) / l,
where k is a proportionality constant. We are given that a beam with a width of 6 inches, depth of 24 inches, and length of 6 inches can safely support 6,336 pounds.
Hence, we can write
6,336 = k * (6 * 24^2) / 6.
Simplifying, we find
k = 6,336 / (6 * 24^2).
To find the load that a beam of 18 inches length can support, we can set l = 18 and solve for p while the width and depth remain constant.
Substituting the known values in the equation gives us
p = (6,336 / (6 * 24^2)) * (6 * 24^2) / 18,
hence p = 6,336 / 18 = 352 pounds.
Therefore, the safe load that the beam can support if it is 18 inches long is 352 pounds.
