Tension of two wires at 40 and 45 degrees

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Homework Help Overview

The problem involves determining the tension in two wires supporting a 40 kg advertising sign, with the wires positioned at angles of 40 and 45 degrees. The original poster expresses uncertainty about the relevant formulas and how to approach the problem.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the importance of drawing a free body diagram and listing force equations in both x and y directions. There are attempts to establish relationships among the forces to solve for unknown tensions.

Discussion Status

Some participants have provided guidance on setting up equations based on the forces acting on the sign. There is a recognition of the need to balance forces for equilibrium, and some participants are exploring simultaneous equations to find the tensions in the wires.

Contextual Notes

Participants mention using gravitational acceleration as 10 m/s² and discuss the assumption that the total upward force must equal the downward gravitational force for equilibrium. There is also a clarification regarding the notation used in force equations.

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;

[tex]T_1\cos40 + T_2\cos45 = 400[/tex]

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?
[tex] \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} & {\cos 45} & {400} \\<br /> {\sin 40} & {\sin 45} & 0 \\<br /> \end{array}} \right] \\ <br /> \left| {T_1 } \right| = 354.9408\,N \\ <br /> T_2 = 310.8120\,N \\ <br /> \end{array}[/tex]

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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