# Tension of two wires at 40 and 45 degrees

In summary, the conversation discusses how to find the tension in each wire holding up a suspended advertising sign with a mass of 40kg. The first step is to draw a free body diagram and list the relevant force equations in the x and y direction. It is important to consider the angles of the rope and use trigonometric functions. By solving the equations simultaneously, the tension in each wire can be determined to be approximately 354.9408N and 310.8120N. The conversation also includes a clarification of a mathematical error.

## 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 [Broken]

I have no idea how to solve this, or what formulas a re relavent to it. Any suggestions are greatly appreciated.

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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?
$$\begin{array}{l} T_1 \cos 40 + T_2 \cos 45 = 400 \\ T_1 \sin 40 + T_2 \sin 45 = 0 \\ \left[ {\begin{array}{*{20}c} {\cos 40} & {\cos 45} & {400} \\ {\sin 40} & {\sin 45} & 0 \\ \end{array}} \right] \\ \left| {T_1 } \right| = 354.9408\,N \\ T_2 = 310.8120\,N \\ \end{array}$$

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

## 1. What is the formula for calculating the tension of two wires at 40 and 45 degrees?

The formula for calculating the tension of two wires at different angles is T = W / sinθ, where T is the tension, W is the weight of the object being supported by the wires, and θ is the angle between the wires.

## 2. How does the angle between the wires affect the tension?

The angle between the wires affects the tension because as the angle increases, the tension decreases. This is because a larger angle results in a smaller value for sinθ in the tension formula.

## 3. Can the tension of the wires be greater than the weight of the object?

Yes, it is possible for the tension of the wires to be greater than the weight of the object. This can occur when the angle between the wires is small, resulting in a larger value for sinθ in the tension formula.

## 4. What happens to the tension if one of the wires is longer than the other?

If one of the wires is longer than the other, the tension in the longer wire will be greater. This is because the weight of the object is distributed over a longer length in the longer wire, resulting in a larger value for W in the tension formula.

## 5. Is there a maximum angle at which the wires can be placed without breaking?

Yes, there is a maximum angle at which the wires can be placed without breaking. This angle is 90 degrees, as any angle greater than 90 degrees would result in a negative tension value, which is not physically possible.