# Joint variation

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• Apr 23rd 2013, 12:37 AM
alex123
Joint variation + percent
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?

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

thank you.
• Apr 23rd 2013, 01:46 AM
agentmulder
Re: Joint variation
The setup is

(1) $W = \frac{kd^2}{l}$

where k is the constant of variation.

The percentage increase of d is $\frac{x}{100} \cdot d$ W remains the same but l becomes 2l.

$W = \frac{k(d + \frac{xd}{100})^2}{2l}$

(2) $W = \frac{kd^2(1 + \frac{x}{100})^2}{2l}$

Set (1) = (2) and solve for x

$\frac{kd^2}{l} = \frac{kd^2(1 + \frac{x}{100})^2}{2l}$

We see k and l and d^2 cancel immediately giving

$2 = (1 + \frac{x}{100})^2$

Can you continue from here?

I'm already estimating x about 41 %...

:)
• Apr 23rd 2013, 04:15 AM
alex123
Re: Joint variation
Quote:

Originally Posted by agentmulder
The setup is

(1) $W = \frac{kd^2}{l}$

where k is the constant of variation.

The percentage increase of d is $\frac{x}{100} \cdot d$ W remains the same but l becomes 2l.

$W = \frac{k(d + \frac{xd}{100})^2}{2l}$

(2) $W = \frac{kd^2(1 + \frac{x}{100})^2}{2l}$

Set (1) = (2) and solve for x

$\frac{kd^2}{l} = \frac{kd^2(1 + \frac{x}{100})^2}{2l}$

We see k and l and d^2 cancel immediately giving

$2 = (1 + \frac{x}{100})^2$

Can you continue from here?

I'm already estimating x about 41 %...

:)

Really appreciate it. Thank you :)
• Apr 23rd 2013, 05:18 AM
agentmulder
Re: Joint variation
Quote:

Originally Posted by alex123
Really appreciate it. Thank you :)

You're welcome, if you get stuck trying to do part b post again.

:)