SOLUTION: The safe load LF a wooden beams to put it up both and varies jointly as the wet W in the square of the deputy and inversely as the length , l . a wooden beams 7 inches wide 9 inche

Question 1186499: The safe load LF a wooden beams to put it up both and varies jointly as the wet W in the square of the deputy and inversely as the length , l . a wooden beams 7 inches wide 9 inches deep and 7 feet long holds up 39701 pounds what load will they be 9 inches wide 6 inches deep and 18 foot long of the same material support round your answer to the nearest integer if necessary
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52776) (Show Source): You can put this solution on YOUR website!
.

Soup of words.

Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!

When you post a problem, politeness should be shown if you want someone to respond. You are not posting a message to a friend on your favorite social messaging network.

Type your message using proper English, including punctuation. And READ what you are posting before you post it.

In this problem, the maximum load varies directly as the width w, directly as the square of the depth d, and inversely as the length l. The given information is that the maximum load is 39701 pounds when the width is 7 inches, the depth is 9 inches, and the length is 7 feet. We are asked to find the maximum load when the width is 9 inches, the depth is 6 inches, and the length is 18 feet.

One way to solve the problem is to use the variation formula with the given set of data to find the constant of proportionality using the given information, then use that constant with the new set of data to find the new safe load.

But since we are only asked to find the safe load for one other set of data, we don't need to do that much work. Instead, we can just determine how the maximum load changes for the change in each of the parameters.

The width changes from 7 to 9 inches. Since the safe load varies directly with the width, this changes the safe load by a factor of (9/7).
The depth changes from 9 to 6 inches. Since the safe load varies directly as the square of the depth, this changes the safe load by a factor of (6/9)^2.
The length changes from 7 feet to 18 feet. Since the safe load varies inversely as the length, this changes the safe load by a factor of (7/18).

So the safe load with the new parameters is

= 8822.4444...

Rounded to the nearest integer, the new safe load is

ANSWER: 8822 pounds

RELATED QUESTIONS
The safe load, L, of a wooden beam supported at both ends varies jointly as the width, w, (answered by josgarithmetic,greenestamps)
The safe load, L, of a wooden beam supported at both ends varies jointly as the width, w, (answered by josgarithmetic)
In problems 11 and 12, use the fact that the load a beam with a rectangular cross section (answered by Mathtut)
The safe load a beam can support varies jointly as the width and the square of the depth... (answered by Earlsdon)
In problems 11 and 12, use the fact that the load a beam with a rectangular cross... (answered by stanbon)
In problems 11 and 12, use the fact that the load a beam with a rectangular cross... (answered by solver91311)
The crushing load of a square wooden post varies directly as the fourth power of it's... (answered by stanbon)
The drag force F on a boat varies jointly with the wet surface area A of the boat (answered by josgarithmetic)
The load that can supported by a rectangular beam varies jointly as the width of the beam (answered by onlinepsa,ikleyn)