Question

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, I.A wooden beam 7 in. wide, 6 in. deep, and 19 ft long holds up 5075 lb. What load would a beam 4 in. wide, 9 in. deep and 14 ft long of thesame material support? (Round off your answer to the nearest pound.)Answer How to enter your answer (opens in new window) 7 PointsKeypadKeyboard ShortcutsIb

Answer 1

Answer: 8855 lb

Explanation

• The joint variation is a type of variation in which the value of a variable depends on two or more variables.

,

• The inverse variation is a type of variation where the value of one variable increases while the value of other variable decreases.

As we are said that L varies jointly as w and the square of d, an algebraic expression of this is:

$L=k\cdot w\cdot d^2$

where k is some non-zero constant.

Additionally, we are said that the latter expression varies inversely as l:

$L=\frac{kwd^2}{l}$

We are given one situation where w = 7 in, d = 6 in, l = 19 ft, and L = 5075 lb.

First, we have to homogenize all the units by changing the length from ft to in:

$19ft\cdot\frac{12in}{1ft}=228in$

Then, we can get the value of k:

$5075=k\frac{7\cdot6^2}{228}$

Simplifying:

$5,075=k\frac{21}{19}$$k\approx4591.67$

Finally, we are told to find L when w = 4 in, d = 9 in, and l = 14 ft. Again we have to calculate the length in inches:

$14ft\cdot\frac{12in}{1ft}=168in$

Then, calculating L with the new values:

$L=4591.67\times\frac{4\cdot9^2}{14}$

Simplifying:

$L\approx8855.36lb\approx8855lb$