Another tension force question

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Homework Help Overview

The discussion revolves around a tension force problem involving a wire, where the tension is given as 13.38 N and the angles are specified as 14 degrees below the x-axis. Participants are trying to determine how to correctly apply trigonometric functions to resolve the tension into its components.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of Tcos(theta) and Tsin(theta) to find the components of tension but express uncertainty about which angle to use. There are questions about the forces acting in the x and y directions, particularly regarding the absence of acceleration and the implications for the normal force.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations of the problem. Some have suggested that the forces in the x direction sum to zero, while others are focused on clarifying the correct angle for the y components. Guidance has been offered regarding the direction of the components, but no consensus has been reached on the angle to use.

Contextual Notes

Participants note the lack of mass and acceleration information, which may affect their calculations. The angles are described as being below the x-axis, leading to questions about how to measure them in the context of the equations being used.

cbchapm2
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Homework Statement


See the attachment for the diagram and problem:
Ch4C WA2.jpg



Homework Equations


So the magnitude of the tension in the wire is 13.38 N. And each angle is 14 degrees below the x-axis, one on the positive side and one on the negative side.
Would I use my Tcos(theta) and Tsin(theta) equations for this? If so, which angle do I use? Because it would be counter clockwise from the x-axis, so NOT 14*, and they wouldn't be the same angle either.


The Attempt at a Solution


I don't really know how to start; I think I have everything I need, I just don't know which step to take first.
 
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Well ƩF=ma, so what are the forces acting on this?
 
I don't have a mass or acceleration though...is there a normal force exerted on it?
 
Ok, how about this, what are the forces in the x and y direction? There is no acceleration, therefore a=0 right? You have to find the force in each direction set them equal to 0 and you'll have the tension. ƩTx=what? ƩTy=what?
 
I think I got that part...
would you find the x and y components using Tcos(theta) and Tsin(theta) for each one? Do you have to do it for each angle? For the angle, would you do counter clockwise from the x-axis?
 
Well since the tensions are the same, it'd just be 2Tsinx for the y direction right? What happens to the x direction?
 
What angle would you use though?

And I'm not really sure for the x direction...is it 0? Or is it 2Tcos(theta)?
 
Ok this about the x direction. One Tx is going in the negative and one is going in the positive. If you sum them up, they become Tcosθ-Tcosθ and that equals 0 (because force is a vector, so it has a magnitude and direction, one positive and the other negative). So the forces in the x direction are 0. The y direction is different though.
Think about the direction of the y components.
 
The sum of the y components would be the 2Tsin(theta). I really just don't know what angle to use for theta.
 
  • #10
First off, what is the direction of the y component.
Second, the degree is in the picture? 14 degrees
 
  • #11
Is the degree in the picture the correct one to use in the equation? Since you have to put in the x and y coordinates; it might be 14 degrees BELOW the x-axis on either side, but the angle is measured counter-clockwise from the positive x-axis. So that's why I'm questioning it. I have the magnitude to plug into the 2Tsin(theta) equation, I just need an angle.
 

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