# Homework Help: Newton's Laws

1. Oct 15, 2008

### Norway

1. The problem statement, all variables and given/known data
Two bricks with mass = 1,0 kg and 1,5 kg is tied together with a thin line going through two rings (with no friction). First, we hold the heaviest brick up so that the two bricks hang still at equal height. Then we'll remove our hand so that the bricks start moving. What is the velocity of the bricks when the lightest brick is 0,26 m higher than the heaviest brick?

Tried to illustrate it, but I'm no good with Paint. But hey, I tried :) Hope you get something out of it.

2. Relevant equations
Don't know.
F=ma
G=mg
Newtons Laws

3. The attempt at a solution
I'm sorry, but I've just sit here all night and stared at this task. I just have no clue. I know you don't like helping people who can't show their work, but I hope someone could still help me. I'm clueless.

Thank you very, very much!

Btw, the answers supposed to be 0,71 m/s.

2. Oct 15, 2008

### Hootenanny

Staff Emeritus
A good way to start would be to consider the forces acting on each brick and then apply Newton's second law to each brick individually.

3. Oct 15, 2008

### mgb_phys

Ignore the pulley for now and just imagine a falling block.
What do you know about the accelaration of a falling object?
What equation links time / speed / accelaration ?

4. Oct 15, 2008

### Norway

Wow, thanks for the quick replies!
I'll try and have another look on it as you said, Hootenanny :)
mgb_phys; well, the acceleration is 9,81 m/s^2, right? And its Force is mass * acceleration. I'll try to use that. Meanwhile, thanks for any further help! :D
As for that equation, do you mean v=v0+at?

Thanks a lot guys :)

Last edited: Oct 15, 2008
5. Oct 16, 2008

### Norway

Well, I couldn't do it. At first I tried a method which returned the answer v = 0,92 m/s which obviously was wrong. Then I tried another approach, but that returned v=1,6 m/s, which also was wrong, obviously.

So, are there any other who can give me some hints?

6. Oct 16, 2008

### Hootenanny

Staff Emeritus
What did you actually try?

7. Oct 16, 2008

### Norway

Sorry, didn't have much time when I posted the previous post, but this is what I tried:
(M is the brick with 1,5kg, m is the brick with 1,0kg)

GM = 1,5 kg * 9,81 m/s2 = 14,715 N
Gm = 1.0 kg * 9,81 m/s2 = 9,81 N

Looking at M (wow, just realised you support TeX :D ):
$$\Sigma F = Ma \Rightarrow a = \frac{\Sigma F}{M} = \frac{4,905 N}{1,5 kg} = 3,27 m /s^2$$

Wrongly assumed that the acceleration was 3,27 metres per second squared.

$$v^2 = v_0^2 + 2as$$
$$v = \sqrt{2 \cdot 3,27 m/s^2 \cdot 0,13 m} = 0,92 m/s$$
Which was wrong, of course. (Kinda embarrassing writing down your work when you know it's wrong, but.. :D )

Tried again: Can't remember what I was thinking, but obviously something very wrong, as this turned out very bad.
$$G_M - S = Ma$$
$$G_m - S = ma$$
$$S = G_m - ma = 9,81 N$$

$$G_M - G_m + ma = Ma$$
$$G_M - G_m = a(M-m)$$
$$a = \frac{G_M - G_m}{M - m} = \frac{4,905 N}{0,5 kg} = 9,81 m/s^2$$
which obviously was even worse. Continued with it even though, and got v = 1,6 m/s.

Well.. - yeah. :)

8. Oct 16, 2008

### Norway

Sorry, but somethings gone wrong here. I can't edit posts, so I posted a new one and can't delete the old one. Sorry. But I split the lines in the last post, so it's easier to see. Thanks

9. Oct 16, 2008

### Hootenanny

Staff Emeritus
Okay, you've got the right idea but are making a few small slips along the way. Let's do this step by step.

Intuitively, in which direction will the system move? In other words, which block will move upwards and which block will move downwards? We shall take this direction as positive.

Now write down the sum of the forces acting on each block, taking note of the direction. If the force acts in the same direction as the block is travelling then the force should be positive. Otherwise, the force should be negative.

Do you follow?

P.S. I've deleted your previous post. You can edit or delete your own posts (up to one hour after to posted them) by clicking the edit button and then either editing the post directly, or selecting the delete option.