Yet Another Simple Tension Problem

  • Thread starter wetcarpet
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In summary, you need to consider both angled strings at the same time, setting up their net forces in both the x and the y directions. You will then get a system of equations that you can solve for the individual tensions.
  • #1
wetcarpet
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A 198N bird feeder is supported by three cables. Find the tension in each cable.
For the angles, see this picture: http://www.webassign.net/sf5/p4_13.gif

{a} Yet again, I split the larger triangle in two by using a positive 198N force, making the large triangle's lower 90 degree angle into a 30 degree angle on the left, and a 60 degree angle on the right.

{b1} Thus, for the tension in the left string:
sin(60) = 198N/Hypotenuse
Hyp. = 198N/sin(60)
Hyp. = 228.6N
--Yet, this value is not the answer, and dividing by two does not work in this instance.
{b2} And for the tension in the right string:
cos(30) = 198N/Hyp.
Hyp. = 198N/sin(30)
Hyp. = 396N
-- Again, this is not the right answer, and divding it by two does not work.

What am I doing wrong?
 
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  • #2
You need to consider both angled strings at the same time, setting up their net forces in both the x and the y directions. You will them get a system of equations that you can solve for the individual tensions.
 
  • #3
The reason dividing by two worked in your other post is because the angles were equal, they shared the weight of the hanging object. In this problem you will find that the strings have different tensions.
 
  • #4
Okay, I tried your method to the best of my ability, and the answer I got was 457.3N for the right string. Is that the correct answer? And if not, could you perhaps explain the process better to me so that I can retry it? Thank you for replying.
 
  • #5
Ok, let's start with the x direction. The net force in this direction is zero N, so you can set up your first net force equation with this. I will call the string on the right 1, and the one on the left 2

(T1)cos(30)-(T2)cos(60)=0

I used minus because they are in opposite direcion, you could also measure the angle from the same starting position and we wouldn't need to manually put the minus in. Anyways, now we need to figure out an equation for the y axis. We know the net force is 198N, and that all of this force is on the y components of our two strings.

(T1)sin(30)+(T2)sin(60)=198

Solving these for T1 and T2 should give you the correct answers for the top two, the tension in the third string should be obvious.
 
  • #6
Okay, I follow your logic thus far, and it makes perfect sense. My only problem is that I am not sure how to solve those equations without at least one value plugged in for T1 or T2. Again, I'm sure it's something right beneath my nose that I'm neglecting here. :smile:
 
  • #7
Substitution, solve one of the equations for one of your unknowns, say T1

T1cos(30)-T2cos(60)=0

T1cos(30)=T2cos(60)

T1=(T2cos(60))/cos(30)

Substituting this into the other equation will allow you to find T2, then to solve for T1 simply plug the value you found for T2 into the above equation.
 
  • #8
Again, I follow your logic perfectly, but another difficulty has cropped up. How on Earth do I isolate T2 in this monster?

((T2)cos60/cos30)sin30+(T2)sin60 = 198

I have made several attempts at this already, and all were failures- forcing me to realize that I need to brush up on some Algebra, hehe.
 
Last edited:
  • #9
It is easiest if you simply throw all those evil trig values into your calculator, or you could recall that

sin(30)=0.5
cos(60)=0.5
cos(30)=Sqrt(3)/2
sin(60)=Sqrt(3)/2

The equation simplifies to
[tex] \frac {0.5}{\sqrt{3}}T_{2}+\frac{\sqrt{3}}{2}T_{2}=198[/tex]
[tex] \frac{4\sqrt{3}}{6}T_{2}=198[/tex]
[tex]T_{2}=\frac{297}{\sqrt{3}}[/tex]
or [tex] T_2\approx171.473[/tex]
 
  • #10
I sincerely thank you for your time and patience. :smile:
 
  • #11
No problem, it was either
a) help you, rip my cds onto my computer, and read my book
or
b) write my english paper that is due tomorrow

and I really didn't want to write that paper. Well, now it is 3:30 in the morning and I have gotten about half of it done. This is really not a good way to start my week...
 

What is "Yet Another Simple Tension Problem"?

"Yet Another Simple Tension Problem" is a hypothetical scenario used to demonstrate the concept of tension in physics and engineering. It involves a rope or cable being pulled at both ends with a certain force, and the resulting tension or stress on the rope is calculated using mathematical equations.

What is tension?

Tension is a force that is transmitted through a medium, such as a rope or cable, by pulling on both ends. It is measured in units of force, such as newtons or pounds, and is typically represented by the symbol "T".

Why is tension important?

Tension is important in many fields, including physics, engineering, and construction. It is a crucial factor in determining the strength and stability of structures, such as bridges and buildings. Understanding tension is also important in designing and maintaining machines and other mechanical systems.

How is tension calculated?

The equation for calculating tension is T = F * L, where T is the tension, F is the force being applied, and L is the length of the rope or cable. This equation assumes that the rope is in a state of equilibrium, meaning that the forces acting on it are balanced.

What factors can affect tension?

The tension on a rope or cable can be affected by a variety of factors, including the magnitude and direction of the forces being applied, the length and elasticity of the rope, and any external factors such as wind or friction. Additionally, the shape and weight of the object being supported by the rope can also impact the tension.

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