Subing, simplifying, isolating and solving Algebra

The book I am reading has me doing the following.

Ft, M and V^2 are unknown values. I am trying to find V^2 by substituting one equation into the next, canceling the M's and solving for V^2. I might be making a couple of mistakes. I need to know what those mistakes are. Thanks.

Solve for Ft.
(Ft)(sin15)=(M)(V^2/R)

(Ft)=(M)(V^2/R)/sin15

Substitute Ft above into Ft below.
(Ft)(cos15)=(M)(G)

(M)((V^2/R)/sin15)(cos15)=(M)(G)

The M's cancel after division, I think.
(M)((V^2/R)/sin15)(cos15)/M=(M)(G)/M

((V^2/R)/sin15)(cos15)=(G)

Multiply V^2 on left to move it to the right.
((V^2)(V^2/R)/sin15)(cos15)=(G)(V^2)

(R/sin15)(cos15)=(G)(V^2)

Divide G on the right to move it to the left.
(R/sin15)(cos15)/G=(G)(V^2)/G

Now V^2 is isolated.
(R/sin15)(cos15)/G=V^2
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kadinAsked:
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HainKurtSr. System AnalystCommented:
maybe after this step:

(M)((V^2/R)/sin15)(cos15)=(M)(G)
V^2 Cos15 / (R * sin15) = G
V^2 (Cos15/Sin15)= GR
V^2 = GR* (Sin15 / Cos15) = GR Tan15
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kadinAuthor Commented:
Thanks for your response. I studied your example but am having a little difficulty understanding it.

Do the M's disappear the same way I divided them away?

I must say I have something to learn about how to get rid of the R. It is confusing to see a fraction inside of another fraction. It looks like you multiplied it by sin15  

V^2 Cos15 / (R * sin15) = G

and that made the R disappear from the left. and of course it's multiplied on the right, that part makes since to me.

It looks like you multiplied (Cos15/Sin15) to move it to the right. Shouldn't that be divided to the right? Thanks.

(M)((V^2/R)/sin15)(cos15)=(M)(G)
V^2 Cos15 / (R * sin15) = G
V^2 (Cos15/Sin15)= GR
V^2 = GR* (Sin15 / Cos15) = GR Tan15
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abbrightCommented:
((V^2/R)/sin15)(cos15)=(G)

Multiply V^2 on left to move it to the right.
((V^2)(V^2/R)/sin15)(cos15)=(G)(V^2)

I think your mistake is this step.
Multiplying things with V^2 doesn't cancel it on the left side, it rather gives

((V^4)/R/sin15)(cos15)=(G)(V^2)

If you want to isolate V^2 you can do it like this:
That's the start:
((V^2/R)/sin15)(cos15)=(G)

Simplify things a bit
(V^2*cos15)/(R*sin15)=G

Divide by cos15:
V^2/(R*sin15)=G/cos15

Multiply by R*sin15:
V^2=(G*R*sin15)/cos15
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BigRatCommented:
The simplest way to do this sort of thing is to divide one equation by the other. That means divide the left hand side of the first equation by the left hand side of the other, and the same for the right hand sides. You'll see immediately that the left hand sides divide out the Ft and sin15/cos15 remain. This is of course tan15. The right hand sides both contain m, so that goes leaving v²/rg, so we then have tan15=v²/rg, so swopping the rg to the other side v²=rgtan15.

This trick is applicable when equations contain the same things (which are to be eliminated) on the same sides. If you try it conventionaly then you'd define Ft either from the first equation, by moving the sin15 to the denominator on the other side, and then place this rather cumbersome term into the left hand side of the second equation and then start eliminating. Alternatively one uses the second equation by moving the cos15 to the other side and substituting the Ft into the left hand side of the first equation. Less cumbersome as before, but still a lot of work. Dividing the two equations is far easier, since Ft divides into Ft and m into m.

HTH
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thehagmanCommented:
Another possible way, though possibly not as obvious:
Divide the first equation by m sin 15°, thus F_t / m = v^2 / ( r sin 15°)
Divide the second equation by m cos 15°, thus F_t / m = g / cos 15°.
You have two equations F_t / m = ..., hence you can equate the right hand sides:
v^2 / ( r sin 15°) =  g / cos 15°
Multiplication with r sin 15° preoduces
v^2  = r sin 15° g / cos 15°
(and once again use sin 15° / cos 15° = tan 15°)
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kadinAuthor Commented:
Thank you all. All of your solutions help me learn. I wish I knew of a book that really covers this specific issue. My current book Elementary Technical Mathematics doesn't explain the more deeply complicated scenarios such as this problem. I have looked at engineering math books at Amazon but I don't want to spend $50 without being certain it will cover all or most of these more complicated scenarios.
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