# Homework Help: Really Simple Acceleration Question

1. Sep 11, 2010

### mohabitar

This should be real easy to figure out, but for some reason it's not coming to me. I'm looking through my notes for similar examples or equations, but cant seem to find anything relative. Since the first block is heavier than the second one, the first one shouldnt move at all. But since theres no friction, then it does move? Or no? If it does move, then would its acceleration be equal to that of block 2? Not really sure where to start.

2. Sep 11, 2010

### rl.bhat

M2 will not fall freely. So for M2

M2*g -T = m2*a.

Since M1 and M2 are connected by a string, both must have the same acceleration. Tension T causes the acceleration in M1.

Hence M1*a = T.

Solve the equations to find a.

3. Sep 11, 2010

### mohabitar

I'm sorry, still a little confused here. So both equations should give me the same answer? Also, wouldnt I need 'a' to find tension? If not, how exactly would I find it?

4. Sep 11, 2010

### collinsmark

Yes it moves! :tongue2:

Without any friction, or any other forces (besides the tension), there is nothing to hold it back.
That's right, well in terms of magnitude anyway. Block 1 will accelerate to the right and Block 2 down. The directions are different but the magnitudes are the same. You can assume the string attaching the two masses does not stretch. So the magnitudes of their velocities are the same, and the magnitudes of their accelerations are the same.
A free body diagram is always a good place to start. Draw all the force vectors acting on each block.

A key to solving this problem and (many like it) is that a massless and frictionless pulley does not affect the tension on the string. The tensions on both sections of the string are the same.

Beyond that, I can't be of any more help until you show us your attempted solution.

[Edit: Yup, rl.bhat beat me to the post yet again. ]

5. Sep 11, 2010

### mohabitar

Ok so the question is dont I need a to find t? Im sure there is no specific equation for finding tension, so how would I go about finding tension?

6. Sep 11, 2010

### mohabitar

Ok well I figured out how to find tension, found something online.
I got the right answer, but I dont understand how the equation works.

T=(m[1] x m[2] x [g]) / (m[1] + m[2])

How do they figure?

7. Sep 11, 2010

### PhanthomJay

It's one thing to look up an equation, another to understand it, so it is a good thing that you are asking for understanding. Rl.bhat and collinsmark already gave you the method. Solve the 2 simultaneous equations with 2 unknowns to find T and a.

8. Sep 11, 2010

### mohabitar

Well right I already got the right answers for both T and A. A=3.3 and T=20.79. But the question I have now is for the T equation. I would have never figured that out had I not looked it up. Whats the reasoning behind that equation? Why would I multiply both masses and divide by their sum and then multiply that whole thing by gravity?

9. Sep 11, 2010

### PhanthomJay

it's the solution of the 2 simultaneous equations, using letters instead of numerical values. It's harder to solve simultaneous equations using letters instead of numbers. That's probably why you shouldn't have looked it up. Instead, use rl.bhat's equations (assuming you understand how they were derived from newton's 2nd law):

m2*g -T = m2*a.
T = m1*a

m2(g) = (m2 + m1)a
32 = 9.5 a
a = 3.3

now since T = m1(a), then T = 20.8

This is just algebra. Which is important to know. But it is more important to understand Newton's laws and free body diagrams. Never blindly look up equations that seem to fit the bill.

10. Sep 11, 2010

### mohabitar

Ohhh ok I see I was supposed to put T from the 2nd equation into the first one. I missed that part. So speaking about this equation m2*g -T = m2*a, why does it work?

We see that the mass of the 2nd block x gravity minus the tension of the string = mass of 2nd block x acceleration. I dont understand the logic behind it. And we dont really use textbooks in this class, its all online and to find out the intuition behind a certain equation online is kind of hard, so any help here would be greatly appreciated.

11. Sep 12, 2010

### collinsmark

The general idea is Newton's second law. There are a number of ways to phrase Newton's second law, but here is a common one*:

The mass m of a given body times the acceleration a of that body equals the sum of all forces acting on that particular body.

$$m \vec a = \sum_i \vec F_i$$

*(This version of the equation assumes constant mass.)

So what does all that really mean? Well, first draw a free body diagram. Draw all the force vectors onto the diagram. For a particular body, add up all the forces acting on that particular body (make sure you do a vector sum). The result is equal to that body's mass times its acceleration.

So let's look at block 2. There are two forces acting on it: The force of gravity mg in the down direction and the tension T in the up direction. If we define down as positive, then the sum of all forces acting on Block 2 is:

$$\sum_i \vec F_i = mg - T$$

Set that equal to m2a.

$$m_2a = m_2g - T$$

Newton's second law in action! http://www.websmileys.com/sm/happy/535.gif Of course we don't know what T is yet, but we'll come back to that.

There is only one force acting on Block 1, and that is the tension T.

$$m_1a = T$$

Since the pulley is massless and frictionless, we know that the two tensions are equal. So combining our equations

$$m_1a = T = m_2g - m_2a$$

$$m_1a = m_2g - m_2a$$

and solving for a,

$$m_1a + m_2a = m_2g$$

$$= a(m_1 + m_2) = m_2g$$

$$a = \frac{m_2g}{m_1 + m_2}$$

So now, minimize your web browser (you can come back to it later if you have to), take out your pencil and paper and repeat what I just did. But please don't try to look back and memorize what I just did, that defeats the point. Instead just remember the following:

(1) Draw your free body diagram. Doing so is really useful.
(2) Newton's second law: ma = sum of all the forces acting on that particular body.
(3) You might not be able to find the final answer for a by analyzing just one body. But that's okay, move on to a different body and combine equations later.

Pencil, paper, go!

12. Sep 12, 2010

### mohabitar

Wow thanks a lot for that! Didnt expect that :) So many things to remember, but I wish I can somehow get to the point where this is all common sense to me, which it appears it is-just common sense. I guess I'll just have to practice a lot, but time is a huge issue. Anyway thanks again!