Equilibrium problem: Calculate the tension of each cord

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Homework Help Overview

The discussion revolves around a problem related to calculating the tension in cords within an equilibrium context. Participants are exploring the relationships between forces and angles involved in the setup.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various mathematical expressions related to tension and angles, including the use of trigonometric identities. Some suggest drawing a force triangle and applying the sine rule as a potentially simpler method. Questions about the correctness of specific calculations and the importance of units in final answers are also raised.

Discussion Status

The discussion is active, with participants providing insights and alternative methods for approaching the problem. There is an exchange of ideas regarding the force triangle and its utility in solving the problem, although no consensus has been reached on a single method or solution.

Contextual Notes

Participants note the importance of including units in answers and the potential for errors in calculations. The discussion reflects a learning environment where different methods and assumptions are being examined without definitive conclusions.

TheePhysicsStudent
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Homework Statement
Was practising an equilibrium problem (and i have done quite a few like this one before and got them right) and I am unsure Where i have went wrong here
Relevant Equations
t1v + t2v = 2.8
1706958890868.png
The question
1706958924448.png
What I did
1706958951569.png
The answer
 
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##\frac{\cos\theta}{\sin\theta}=?##
 
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Not ##\tan## :smile:
(slow typist)
 
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Also, $$\begin{align} & T_2\frac{\cos60^{\circ}\cos40^{\circ}}{\sin40^{\circ}}+
T_2\sin60^{\circ}=T_2\left(\frac{\cos60^{\circ}\cos40^{\circ}+\sin60^{\circ}\sin40^{\circ}}{\sin40^{\circ}}\right) \nonumber \\
& =T_2\frac{\cos(60^{\circ}-40^{\circ})}{\sin40^{\circ}} =T_2\frac{\cos20^{\circ}}{\sin40^{\circ}}.\nonumber
\end{align}$$
 
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TheePhysicsStudent said:
A couple of things worth noting...

1. If you draw the force-triangle and use the sine rule, the problem takes only a few lines of simple working. (It’s a lot quicker and less error-prone than the method you used.)

2. You have forgotten the unit in your final answer - lose one mark!
 
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Hanging toy.jpg
 
To avoid any confusion, by 'force-triangle' (in Post #5) I meant this...
triangle.gif
 
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Steve4Physics said:
To avoid any confusion, by 'force-triangle' (in Post #5) I meant this...
View attachment 339715
Wow thanks, I think this information may help speed time in lots of calculations i do
 
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TheePhysicsStudent said:
Wow thanks, I think this information may help speed time in lots of calculations i do
Just in case it is not clear, when drawing a force polygon of forces in balance, the arrows join head to tail, as in @Steve4Physics' drawing.
You can also use them to find the resultant of a system of forces. In that case the resultant completes the polygon but its arrow is reversed.
 
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