# Yet Another Simple Tension Problem

1. Mar 20, 2005

### wetcarpet

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?

2. Mar 20, 2005

### krusty the clown

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. Mar 20, 2005

### krusty the clown

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. Mar 20, 2005

### wetcarpet

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. Mar 20, 2005

### krusty the clown

Ok, lets 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. Mar 20, 2005

### wetcarpet

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.

7. Mar 20, 2005

### krusty the clown

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. Mar 20, 2005

### wetcarpet

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: Mar 20, 2005
9. Mar 20, 2005

### krusty the clown

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
$$\frac {0.5}{\sqrt{3}}T_{2}+\frac{\sqrt{3}}{2}T_{2}=198$$
$$\frac{4\sqrt{3}}{6}T_{2}=198$$
$$T_{2}=\frac{297}{\sqrt{3}}$$
or $$T_2\approx171.473$$

10. Mar 20, 2005

### wetcarpet

I sincerely thank you for your time and patience.

11. Mar 21, 2005

### krusty the clown

No problem, it was either