# Newton Third Law general equation

1. Oct 18, 2006

### bigredd87

I have a picture provided. You can assume that the table is frictionless. I need an expression for the acceleration of m1.

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Last edited: Oct 18, 2006
2. Oct 18, 2006

### HallsofIvy

Staff Emeritus
We don't do your homework for you. What are the forces on each mass?

3. Oct 19, 2006

### bigredd87

here is what I have so far:

Fnetx for m1 is T=m1a
Fnety for m1 is N-m1g=0

I guess my real dilemma is i don't really know what the forces on m2 are. I believe there is no acceleration constraint since m2 is on the second pulley. Does that mean that the tension and weight are equal to m2a, with m2 having a different acceleration than m1?

4. Oct 19, 2006

### Staff: Mentor

OK. (All you need here is Fnetx.)

Treat m2 and its pulley as one object. I see 3 forces acting on it.
Of course there is an acceleration constraint, which you need to figure out. When m2 drops X distance, how far must m1 move?

You will apply Newton's 2nd law to m2, just like to m1. Yes, a2 is different from a1, but there is a simple relationship.

5. Oct 19, 2006

### bigredd87

so a2 =-a1? I didn't understand if the mass on the pulley made it different from other problems. So there is tension T from the rope attached to the wall, the weight force m2g, and the tension from m1 which is m1a1?

6. Oct 19, 2006

### Staff: Mentor

No. (But the minus sign is good.) Answer my question earlier: When m2 drops X distance, how far must m1 move? (Get a piece of string and figure it out.)

The tension will be the same throughout the rope. The only things exerting a force on m2 are: (1) the ropes; (2) gravity.

Set up an equation for each mass. (After you've found the acceleration constraint.) Then solve them together to find the accelerations.

7. Oct 19, 2006

### bigredd87

ok, so I think I got it now.

a2 = (-1/2)a1

so u have for m2 in y direction:
2T-m2g=m2a2 ; since there are two ropes pulling on m2

i already know that T =m1a1 and the acceleration constraint, so plugging this in i get:
2m1a1-m2g=-m2a1/2

4m1a1-2mg=-m2a1

a1(4m1 + m2)= 2mg

a1=2mg/(4m1+m2)
does this look ok?

8. Oct 19, 2006

Excellent!