Joint/Combination Variation: change in F = mnx^2 / p if....

[AngelofContrary]

F = mnx^2 / p

The problem asks what would happen to F if m and n remain constant, x is halved, and p is doubled.

Any ideas?

[n really is the pi sign, but I substituted it for another variable because the sign wouldn't show up.]

 

[tkhunny]

Sure! Plug it in.

m*n*(x/2)^2 / (2p)

Now manipulate to express it in terms of F.

 

[AngelofContrary]

So would I multiply by two to get rid of the two in the denominator, then multiply both sides by four to get rid of the (x/2)^2?

Then I'd come out with 8F = mn(x^2/4) / p..

 

[Denis]

Re: Joint/Combination Variation

F = mnx^2 / p

To see where you are in solving:
can you solve that equation in terms of m?

m = ?

 

[tkhunny]

So would I multiply by two to get rid of the two in the denominator, then multiply both sides by four to get rid of the (x/2)^2?

Then I'd come out with 8F = mn(x^2/4) / p..

What? You sort of have the right idea, but, as you have indicated, that's rather confusing.

\(\displaystyle \L\;\frac{m*n*(\frac{x}{2})^{2}}{2*p}\;=\;\frac{m*n*x^{2}}{2^{2}*2*p}\;=\;\frac{m*n*x^{2}}{p}*\frac{1}{8}\;=\;F*\frac{1}{8}\)