SOLUTION: The maximum load of a horizontal beam that is supported at both ends vari<bes jointly as the width and the square of the height and inversely as the length between the supports. A

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Question 1106811: The maximum load of a horizontal beam that is supported at both ends vari of the height and inversely as the length between the supports. A beam 6 m long, 0.1 m wide, and 0.06 m high
supports a load of 360 kg. What is the maximum load supported by a beam 16 m long, 0.2 m wide, and 0.08 m high?

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The information tells us that the mass m that can be supported by a beam of length l, width w, and height (thickness) h is

m+=+k%28%28w%29%28h%5E2%29%2Fl%29

where k is a constant.

The value of the constant can be determined from the given information: When the length is 6m, the width is 0.1m, and the height is 0.06m, the maximum load is 360kg. So

360+=+k%28%280.1%29%280.06%5E2%29%2F6%29
360+=+k%28.00036%2F6%29+=+k%28.00006%29
k+=+6000000

Then using this constant with the new beam measurements,
m+=+6000000%28%280.2%29%280.08%5E2%29%29%2F16
m+=+6000000%28.00128%2F16%29+=+480

That is a general method for solving problems like this. But if the numbers are fairly simple, as in this case, I find it easier just to modify the given mass according to each changed dimension of the beam.

The length of the beam changes from 6m to 16m; since the mass varies inversely as the length, the maximum load decreases by a factor of 6/16 = 3/8.
The width doubles from 0.1m to 0.2m; since the mass varies directly with the width, the maximum load increases by a factor of 2.
The height changes from 0.06m to 0.08m; since the mass varies directly with the square of the height, the maximum load increases by a factor of (0.08/0.06)^2 = (4/3)^2 = 16/9.

Then the new maximum load is
360%2A%283%2F8%29%2A%282%29%2A%2816%2F9%29+=+360%2A%284%2F3%29+=+480