# Thread: Joint variation + percent question

1. ## 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:

$\displaystyle W \alpha \frac {d}{l}$

$\displaystyle W = \frac {kd}{l}$

$\displaystyle W = \frac {2}{3}$

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?

2. 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:

$\displaystyle W \alpha \frac {d}{l}$

$\displaystyle W = \frac {kd}{l}$

$\displaystyle W = \frac {2}{3}$

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,

$\displaystyle W=\frac{kd}{l}$---------(1)

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

$\displaystyle W=\frac{kD}{2l}$---------(2)

By (1) and (2),

$\displaystyle D=2d$

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

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