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 3 inches wide, 5 inches high, and 120 inches long can hold a load of 2600 pounds. what load would a beam 5 inches wide, 8 inches high, and 168 inches long, of the same material, support?
Load that the beam can support = 7923.81 pounds
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.
where k = proportionality constant
If w = 3 inches, h = 5 inches, l = 120 inches and L = 2600 pounds
k = 4160
If w = 5 inches, h = 8 inches and l = 168 inches
L = 7923.81 pounds
First, get the volume of each and since density is the same, we shall use the density to get the other mass
Volume=lwh where l is length, w is width and h is height.
Substituting with the given figures
Density is mass per unit volume
Mass is given as 7900 pounds
Density is 7900/6048=1.3062169312169
Volume of second box
Mass is product of volume and density hence 1.3062169312169*1728=2257.1428571428 pounds
Rounded off as 2257 pounds
load varies directly as the product of width and square of height and inversely as the length
load = kwh²/ L where w = width in inches, h = height also in inches, L = length in inches and k is constant
2480 = (k × 2 × 7²) / 192
2480 × 192 = k × 2 × 49
k = 476160 / 98 = 4858.78
load that 4 ince wide, 6 inches high, and 120 inches long of the same material support = kwh² / L = 4858.78 × 4 × 6² / 120 = 699664.32 / 120 = 5830.54 pounds
if you divide 10 by 60 the answer is 0.166666 while if you divide or solve the other problems it equals 0.6