The maximum safe load of a rectangle beam varies jointly as its width and the square of its depth and inversely as the length of the beam. If a beam 8.00 in. wide, 11.5 in. Deep, and 16.0 ft long can safely support 15.000 lb, find the safe load for a beam 6.50 in. Wide, 13.4 in. Deep, and 21.0 ft long made of the same material.
I can't figure it out.
To find the safe load for the second beam, we need to use the joint variation relationship given:
The maximum safe load (L) varies jointly as the width (w) and the square of the depth (d^2) and inversely as the length (L) of the beam.
Mathematically, we can write this relationship as:
L = k * w * d^2 / L
Where k is the constant of variation.
To find the value of k, we can use the information given in the problem to solve for it:
For the first beam:
L1 = k * w1 * d1^2 / L1
15,000 = k * 8 * (11.5)^2 / 16
Simplifying the equation, we can solve for k:
k = (15,000 * 16) / (8 * (11.5)^2)
Now that we have the value of k, we can use it to find the safe load for the second beam.
For the second beam:
L2 = k * w2 * d2^2 / L2
L2 = [(15,000 * 16) / (8 * (11.5)^2)] * 6.5 * (13.4)^2 / 21
Calculating the value, we can find the safe load for the second beam.