# I'm stupid

1. Feb 14, 2004

### jjiimmyy101

Situation: You have a 50kg box hanging with two different ropes attached to the box at two different points on the box. (The ropes are hanging vertically)

Question: What would be the tension in each rope?

Answer?: Would you just divide the force of the hanging box (490.5N) into two?

I attached a picture.

#### Attached Files:

• ###### box.gif
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Last edited: Feb 15, 2004
2. Feb 14, 2004

### drag

Greetings !

First of all, no one's stupid.

As for your question, it is ussualy helpfull
in these cases to look at whole systems.
In this case you have just a single body,
but you can treat it as a system - as an example
for more complex cases. Draw the force of gravity - mg.
Now, just draw a circle around the whole system -
just one body in this case. The system is not
moving which means that the total force acting
on it is zero. There are three "lines" - force vectors
exiting your system's circle two up and one down
and their sum is zero. Now practice this a lot
more and you'll no longer have trouble with any of
these hanging bodies/pulleys and strins' Qs.

Live long and prosper.

3. Feb 14, 2004

### ShawnD

You can only divide by 2 if the ropes are equal distance from the centre of mass.

If they are different distances from the centre, you have to make 2 equations - moment and vertical force. The vertical force would just be like this:
F1 + F2 = weight
F1 = weight - F2

Then make the formula for moment:
F1d1 - F2d2 = 0
Then fill in the other equation:
(weight - F2)d1 - F2d2 = 0

From there you can easily rearrange it and solve for F2 then go back to the first equation and solve for F1.

4. Feb 14, 2004

### jjiimmyy101

Thank-you for your kind words and prompt response.

So, if I understand properly, I was correct, but I didn't know why I was correct.

The equation is: 2T - 490.5 = 0

You also mentioned "The system is not moving which means that the total force acting on it is zero." So, if the system is moving, the total force acting on the system is equal to (mass X acceleration)? Am I right?

5. Feb 14, 2004

### jjiimmyy101

So I was wrong. Thanks for the help!

6. Feb 15, 2004

### drag

Well, someone is stupid, but it ain't you guys. [zz)]
Like ShawnD said, if the problem is not symmetric
(and it probably isn't [b(] ) you have
to use the moment of inertia equation too (R1xT1=R2xT2).
Remember it's a cross product and R1,T1,R2 and T2 are
vectors, so you only multiply the distances by the
perpendicuilar components of the forces.

7. Feb 15, 2004

### HallsofIvy

jjiimmyy101: One problem all of us have is that the picture you intended to attach apparently didn't get attached! Try again.

8. Feb 15, 2004

### jjiimmyy101

Thanks for replying everyone!

It's a very simple diagram I drew myself. I'll try to attach it again.

This isn't a question from a textbook. I have a different question that I think involves this principle and I just wanted to understand this to see if I could extend this knowledge to my problem.

#### Attached Files:

• ###### box.gif
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1.6 KB
Views:
109
9. Feb 15, 2004