Finding tension in 2 cords.

[charlie_luna]

Homework Statement

An 80 N traffic light is supported at the midpoint of a line between two poles that are 30 m apart. If the light sags a vertical distance of 1 m at the midpoint, what is the tension in the supporting line?

A 20 lb picture is hung from a nail so that the supporting cords make an angle of 60. what is the tension of each cord segment?

Here is the actual homework problem i have to solve: An 80 N picture is hung from a nail as indicated. Assume the zeros in the 80 N weight are significant. The angle between the cords in 78 degrees. What is the tension in the each cord?

Homework Equations

I am lost here.

The Attempt at a Solution

And lost here.

There are 2 examples in my instructor's lecture notes that he sends to us by mail that has a street light and another picture hanging from 2 cords. in the street light problem, the tensions are all positive:

Tax = TaCosdegree
Tay = TaSindegree
Tbx = TbCosdegree
Tby = TbSindegree

and so on. But in the picture problem, some of the equations have a negative out front. such as:

Tbx = TbCos60
Tby = TbSin60
Tax = -TaCos60
Tay = TaSin60.

why is that? why would some problems be all positive and some get negatives in others? I have realized that i am suppose to set EFx = 0. where the E is the sigma symbol which i believe stands for equilibrium or net. and i am suppose to set EFy = 0.

can someone please help me understand this? my instructor has been out sick since last monday with swine flu and he has not returned to work yet.

 

[Redbelly98]

Welcome to PF.

It's best to start by drawing a free-body diagram. For these problems, there should be 3 forces acting on the object.

 

[tomcue]

I would approach this problem by first identifying the forces acting on the object (the traffic light or picture) and then using the principles of Newton's laws to determine the tensions in the cords.

In the case of the traffic light, we can assume that the weight of the light is acting downwards and is balanced by the tension in the cord pulling upwards. Since the light is sagging, we can also assume that there is a horizontal component of the tension in the cord. Using trigonometry, we can find the magnitude of the tension in the cord by setting the vertical component equal to the weight of the light and solving for the tension.

In the case of the picture, we have a similar situation where the weight of the picture is balanced by the tension in the cords. However, since the cords make an angle with the vertical, we need to take into account the horizontal and vertical components of the tension. Again, using trigonometry and Newton's laws, we can solve for the tension in each cord.

In both cases, it is important to consider the direction of the forces and use appropriate signs for the components of the tension. For example, in the picture problem, the tension in one of the cords may be acting in the opposite direction as the weight of the picture, resulting in a negative sign for that component of the tension.

It is also important to remember to set the net force in both the horizontal and vertical directions equal to zero, as this represents a state of equilibrium. This is what the sigma symbol stands for in your equations.

In summary, as a scientist, I would approach this problem by carefully considering the forces involved and using the principles of Newton's laws and trigonometry to solve for the tensions in the cords. I hope this helps to clarify the process for you.