Tension of two wires at 40 and 45 degrees

AI Thread Summary
The discussion revolves around calculating the tension in two wires supporting a 40 kg sign. Participants emphasize the importance of drawing a free body diagram to visualize the forces and angles involved. The key equations derived from the forces in the x and y directions are T1*cos(40) + T2*cos(45) = 400 N and T1*sin(40) + T2*sin(45) = 0. By solving these equations simultaneously, the tensions in the wires are determined to be approximately 354.94 N for T1 and 310.81 N for T2. The conversation highlights the significance of equilibrium in solving such physics problems.
unique_pavadrin
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Homework Statement


An advertising sign of mass 40kg is suspended by two wires as shown in the diagram. What is the tension in each wire?

http://img132.imageshack.us/img132/9681/untitledwx0.png


I have no idea how to solve this, or what formulas a re relavent to it. Any suggestions are greatly appreciated.
unique_pavadrin
 
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The 1st step, you should do is draw a free body diagram for the hanging sign, that shows the forces acting on it. Include in your free body diagram, the angles of the rope.

2nd step should be listing the relevant Force equations in the x and y direction.
Hint: (Fnet)x = 0 = ...
(Fnet)y = 0 = ...

then try to find a relationship amonst them to solve for the unknown quantities. (Hint: i think you will be better off starting with the (fnet)y equation).
 
okay thanks

so the (Fnet)y=-400 + 400=0
therefore the combined (F)y of the two wire is equal to 400 N?
 
uhh no... how did u get -400+400??
 
i assumed that gravity acceleration is 10 m s-1
then used w=mg to find 400
since the sing pulls down with 400N the two wire must pull up with 400N for the system to be in equilibrium so there 400N acting in both direction for it to cancel each other out
 
You are correct, so you can write;

T_1\cos40 + T_2\cos45 = 400

Now, do the same for the x components.
 
okay thanks Hootenanny
so if i do the following and solve simultaneously, will it yield me the correct result?
<br /> \begin{array}{l}<br /> T_1 \cos 40 + T_2 \cos 45 = 400 \\ <br /> T_1 \sin 40 + T_2 \sin 45 = 0 \\ <br /> \left[ {\begin{array}{*{20}c}<br /> {\cos 40} &amp; {\cos 45} &amp; {400} \\<br /> {\sin 40} &amp; {\sin 45} &amp; 0 \\<br /> \end{array}} \right] \\ <br /> \left| {T_1 } \right| = 354.9408\,N \\ <br /> T_2 = 310.8120\,N \\ <br /> \end{array}<br />

thanks
 
oh i c.. ye that's correct... you wrote (Fnet)y=-400 + 400=0, but it's
Fnet =-400 + 400=0.
 
okay thanks for the help, greatly appreciated
 
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