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Algebra. How to solve with two missing variables?

Posted on 2011-03-24
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Last Modified: 2012-05-11
If you skip to the very bottom of this force problem you find two equations. In a rectangle called step (2) it says; Sub in T3 in symbols; Solve for T2.

The circled variables are unknowns.

How do I solve this when both T2 and T3 are unknowns? By the way, T1 is 25.
This is as far as I get. Thanks.

T2 sin40  +  T3 sin30 =T1
                    T2 sin40 =T1 -  T3 sin30  
                    T2 sin40 =25 -  T3 sin30          


twoD-forces02.jpg
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Question by:kadin
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Assisted Solution

by:aburr
aburr earned 334 total points
ID: 35212830
in x component eq solve for T3

T3 = T2 cos 40/cos30

in y component eq put the above eq in place of T3

T2sin40 - T2 cos 40/cos30 = T1
remember that T1 = 24 N (start of prob). Now the only unknown is T2  - solve
put the T2 in the x eq to get T3

QED

Note that the procedured here is exactly like the solution given you before.
Solve for separate one variable and substitute into the other equations. Repeat if necessary
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Author Comment

by:kadin
ID: 35212851
Thanks. I will need some time to study what you said.
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Author Comment

by:kadin
ID: 35213100
Thanks for your help. I see where I misread the equation. I thought T3  cos30 was one unit when it's actually T3(cos30). Multiply the two. There for I can divide cos30 to the right side thus isolating T3.

My next challenge is how to solve for T2 when there are two T2's in the same equation. How do I isolate two of them?

T2sin40 - T2 cos40/cos30 = T1

Let say I had a simpler equation.

T2(4) - T2(6)/2=24
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LVL 84

Assisted Solution

by:ozo
ozo earned 166 total points
ID: 35213986
T2(4) - T2(6)/2 = T2((4) - (6)/2)
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Accepted Solution

by:
aburr earned 334 total points
ID: 35216067
"T2sin40 - T2 cos40/cos30 = T1"

notice that there is a T2 in both terms on the left. Factor the T@ out

T2(sin40 - cos49/cos30) = T1   that is the same a aT2   divide both sides by a
you get
T2 = T1/(sin 40 - cos40/cos30)
There you have it
I would recommend that you review your simple algebraic operations.
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Author Closing Comment

by:kadin
ID: 35217878
Thanks. I will work on the rest of this and see if I can get the right answer.
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