Solve for the tension in T1 and T2

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SUMMARY

The discussion focuses on solving for the tensions T1 and T2 in a physics problem involving angles and forces. The equations derived include T1sin42 + T2sin18 - 90 = 0 and T1cos42 - T2cos18 = 0. The user initially miscalculated but later corrected their approach, arriving at T2 = 89.23 and T1 = 157.92. The final correction yielded T2 = 65.93, indicating a need for careful consideration of the equations and the hanging weight's force.

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Homework Statement


A friend gave this to me and i have never taken physics could use some help on how to solve this.


Homework Equations


Have no clue


The Attempt at a Solution


Have researched and came up with this probably not correct
T1sin42+T2sin18=0
T1cos42-T2cos18=0
 

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You need to incorporate the force due to the hanging weight into the second equation.

EDIT: Sorry, the *first* equation! The vertical components.
 
Last edited:
lilgen said:

Homework Statement


A friend gave this to me and i have never taken physics could use some help on how to solve this.


Homework Equations


Have no clue


The Attempt at a Solution


Have researched and came up with this probably not correct
T1sin42+T2sin18=0
T1cos42-T2cos18=0

The first equation should be: T1sin42 + T2sin18 - 90 = 0.

Solve the second equation for T1 (in terms of T2). Plug that into the corrected first equation, and solve for T2. Then use that to find T1.
 
T1sin42+T2sin18-90=0
T1cos42-T2cos18=0

T1=T2cos18/cos42

T2cos18/cos42*sin12+sin18-90=0

T2=89.23
T1=157.92

Is this the way you were talking about doing it? I am wanting to learn it thank you for your help.
 
lilgen said:
T1sin42+T2sin18-90=0
T1cos42-T2cos18=0

T1=T2cos18/cos42

T2cos18/cos42*sin12+sin18-90=0

T2=89.23
T1=157.92

Is this the way you were talking about doing it? I am wanting to learn it thank you for your help.

T2(cos18/cos42*sin42+sin18)-90=0

Sorry about not following up.

I get T2=65.93 .
 

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