Newton Third Law general equation

[bigredd87]

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

 

[HallsofIvy]

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

 

[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?

 

[Doc Al]

here is what I have so far:

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

OK. (All you need here is Fnetx.)

I guess my real dilemma is i don't really know what the forces on m2 are.

Treat m2 and its pulley as one object. I see 3 forces acting on it.

I believe there is no acceleration constraint since m2 is on the second pulley.

Of course there is an acceleration constraint, which you need to figure out. When m2 drops X distance, how far must m1 move?

Does that mean that the tension and weight are equal to m2a, with m2 having a different acceleration than m1?

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.

 

[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?

 

[Doc Al]

so a2 =-a1?

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.)

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?

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.

 

[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?

 

[Doc Al]

Excellent!