# This is incredibly hard for Grade 12 homework!

1. Dec 2, 2004

### Catalyst

Incredibly hard Physics homework!

Alright, so here's the problem:

Three known masses are connected together by four pieces of ideal string of known length. Determine the tensions in the strings and the unknown angles, given any value for the length of the strings and the weights.

http://img122.exs.cx/img122/6239/75-physicsproblem.jpg

So far, I've been able to pull this information from the question:

0 degrees < {|A1|, |A2|, |A3|, |A4|} < 90 degrees
The direction that the angles will go will depend on whether the values of the angles are positive or negative.

T1 cos A1 = T2 cos A2 (1)
T2 cos A2 = T3 cos A3 (2)
T3 cos A3 = T4 cos A4 (3)
T1 sin A1 - T2 sin A2 = m1g (4)
T2 sin A2 - T3 sin A3 = m2g (5)
T3 sin A3 - T4 sin A4 = m3g (6)

l1 cos A1 + l2 cos A2 + l3 cos A3 + l4 cos A4 = L (length of top bar) (7)
l1 sin A1 + l2 sin A2 + l3 sin A3 +l4 sin A4 = 0 (8)

Given: l1, l2, l3, l4, L, m1, m2, m3
Find: T1, T2, T3, T4, A1, A2, A3, A4

How does one go about solving these non-linear equations?

I've solved this for 1 mass and halfway done solving this for 2 masses. Solving this for 3 masses is much harder though. Can anyone please help?

Last edited: Dec 2, 2004
2. Dec 2, 2004

### ceptimus

Looks tough. It's quite possible for any one of the four strings to have zero tension in it, and the weights would then all hang directly under one or other of the support points - for example if:

$L1 > \sqrt{(L2 + L3 + L4)^2 + 1}$

then all the weights would hang directly under the right hand support and the L1 string would hang slack.

I would solve this problem with numerical methods - a computer program that initially guesses angles, then calculates the force on each mass, based on the guess, and moves a weight to the right or left if the forces aren't in balance. Repeated iterations of the calculation would bring the out of balance forces arbitrarily close to zero, at which point the solution could be printed. A spreadsheet - e.g. an Excel spreadsheet could do this quite easily.

But I guess that's not the kind of solution you're looking for?

Last edited: Dec 2, 2004
3. Dec 2, 2004

### Distillation

(I'm working with Catalyst on this problem; a group project)
This is the kind of solution we're looking for. We were told to create a program to solve this problem. Hmmm, I think I get what you mean.

4. Dec 3, 2004

### ceptimus

OK. I would start by working out the Cartesian coordinates (x and y) for each mass.

If you have the left hand support as the origin (0, 0) and x increasing to the right, and y increasing vertically downwards (I know this seems backwards, but it stops the y values from all being negative), then m1 is at (x1, y1) where
x1 = L1 cos (theta1) and
y1 is L1 sin(theta1)

now for m2 you get

x2 = x1 + L2 cos(theta2)
y2 = y1 + L2 sin(theta2)

and so on...

By the way, for consistency in labelling the angles, theta4 in your diagram would have a negative value.

If you want some more hints, show me what you've worked out so far.

Last edited: Dec 3, 2004
5. Dec 30, 2004

### aratnaus

I can't solve this problem!

1. A uniform wire has a resistance of 18 ohm. It is bent to the shape of an equilateral triangle. What is the effective resistance between any two corners of the triangle?

6. Dec 31, 2004

### CeeAnne

Resistance Formula

Aratnaus. I believe you should have started a new thread, but since we are here ...
Effectively what you form with the equilateral triangle is three 6 ohm resistors, 18 ohms divided into thirds, so the resistance across any two points would be equivalent to a 6 ohm resistor paralleled with a 12 ohm resistor. You can see what I mean if you sketch the triangle and label each side 6 ohms. The basic formula for the total resistance of two parallel resistors is R1 times R2, divided by R1 plus R2.
Rt = (R1 x R2) / (R1 + R2). You can figure it from this point.

Last edited: Dec 31, 2004
7. Dec 31, 2004

### Diane_

Nicely done, CeeAnne. :)

(I'd've put parentheses around the (R1 + R2), but I'm picky that way.)

8. Dec 31, 2004

### CeeAnne

Thank you, Diane

Thank you, Diane. I did as you suggested.