Thread: Joint variation + percent question

The weight (W) that can be supported by a wooden beam varies directly as the square of its diameter (d) and inversely as its length (l).
a What percentage increase in the diameter would be necessary for a beam twice as long to support an equivalent weight?

What percentage change in the weight would be capable of being supported by a beam
three times as long with twice the diameter?

I know how to do this one:







So the weight needs a 33% increase.


But for this question:
What percentage increase in the diameter would be necessary for a beam twice as long to support an equivalent weight?

How do I do this?

Originally Posted by jgv115

The weight (W) that can be supported by a wooden beam varies directly as the square of its diameter (d) and inversely as its length (l).
a What percentage increase in the diameter would be necessary for a beam twice as long to support an equivalent weight?

What percentage change in the weight would be capable of being supported by a beam
three times as long with twice the diameter?

I know how to do this one:







So the weight needs a 33% increase.


But for this question:
What percentage increase in the diameter would be necessary for a beam twice as long to support an equivalent weight?

How do I do this?

Dear jgv115,

---------(1)

When the beam is twice as long, let the diameter be D that supports the weight W,

---------(2)

By (1) and (2),



So the percentage increase in the diameter would be 100%.