Question

the safe load, l, of a wooden beam of width w, height h, and length l, supported at both ends, varies directly as the product of the width and the square of the height, and inversely as the length. a wooden beam 4 inches wide, 7 inches high, and 240 inches long can hold a load of 2540 pounds. what load would a beam 5 inches wide, 8 inches high, and 216 inches long, of the same material, support? round your answer to the nearest integer if necessary.

Answer 1

The safe load, l, of a wooden beam of width w, height h, and length l, then the nearest integer is, L = 4607.7 pounds.

Based on the given conditions,

The safe load = L

A wooden beam of width = w

Height = h

Length = l

The safe load L, of the beam varies directly as the product of the width and the square of the height.

L∝ (w × h²)

And varies inversely as the length of the wooden beam.

L∝ (w × h²)/l

L = $\frac{k*w*h^{2} }{l}$       (Equation-1)

Where k = proportionality constant

If w = 4 inches, h = 7  inches, l = 240 inches and L = 2540 pounds

Then,

We can substitute values in equation-1,

L = $\frac{k*w*h^{2} }{l}$  

2540 = (k*4*$7^{2}$)/240

2540 = (k*4*49)/240

2540 = (k*196)/240

2540*240 = k*196

609600 = k*196

609600/196 = k

3110.20 = k

If w = 5 inches, h = 8 inches and l = 216 inches

L = $\frac{k*w*h^{2} }{l}$  

L = ( 3110.20 * 5 * $8^{2}$ )/216

L = ( 3110.20 * 5 *64 )/216

L = 995264/216

L = 4607.7 pounds

Therefore,

The safe load, l, of a wooden beam of width w, height h, and length l, then the nearest integer is, L = 4607.7 pounds.

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