Solving Vectors and Forces Homework Problem

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The problem involves calculating the tension in two wires supporting a 150N sign, with one wire at a 50-degree angle and the other at a 40-degree angle to the wall. Initial calculations suggested tensions of 233.36N and 195.81N for the respective wires, but these were incorrect. The correct approach requires setting up equations for both vertical and horizontal components of the forces, ensuring that the sum of the vertical components equals the weight of the sign. A hint was provided to simplify the problem by finding a direction to ignore one of the tension forces. Accurate calculations depend on applying trigonometric principles correctly to resolve the forces.
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Homework Statement



A sign weighs 150n and is suspended on 2 wires between 2 walls, 1 at 50 degrees to the wall and the other at 40 deg to the other wall. What is the tension in each wire?

Homework Equations





The Attempt at a Solution



For the 1st wire F=150/cos50
=233.36N
2nd wire F=150/cos40
=195.81N
Is this likely to be correct I can't find any simular examples to help me out.
 
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Welcome to PF!

Hi P.townsville! Welcome to PF! :smile:

(have a degree: º :wink:)
P.townsville said:
A sign weighs 150n and is suspended on 2 wires between 2 walls, 1 at 50 degrees to the wall and the other at 40 deg to the other wall. What is the tension in each wire?
…
For the 1st wire F=150/cos50
=233.36N
2nd wire F=150/cos40
=195.81N

Nooo :redface: …

you're saying F1cos50º = 150, F2cos40º = 150.

But if you're taking vertical components, that'll be F1cos50º + F2cos40º = 150, won't it?

Hint: you need one equation for both vertical components, and another equation for both horizontal components, and then a bit of algebra and trig. :smile:

(Alternatively, find a direction in which you can ignore one of the tension forces :wink:)
 

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