Calculating acceleration in a pulley system

1. The problem statement, all variables and given/known data
Frictionless, massless, pulley system. Cables attached to m1 are parallel to the tabletop. Cables attached to m2 are parallel to each other, and perpendicular to the ground. I'm trying to solve for the acceleration of the objects in terms of the variables, m1, m2, g, and theta. Also, m1's velocity is up the table and the coefficient of kinetic friction is zero.
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2. Relevant equations
Summation of forces = mass*acceleration (2nd law)

3. The attempt at a solution
So far I've constructed the free body diagrams. I got that the summation of the forces on mass two is equal to 3T - (m2*g). Summation of forces on mass one in the y is 2T(sin(theta))-(m2g). Summation of forces on mass one in the x is 2T(cos(theta)). So I figure a1 = 2/3(a2)? After that I'm just getting lost in a mess of algebra...
 

Doc Al

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I got that the summation of the forces on mass two is equal to 3T - (m2*g).
OK.
Summation of forces on mass one in the y is 2T(sin(theta))-(m2g). Summation of forces on mass one in the x is 2T(cos(theta)).
I advise you to consider components parallel to the incline surface. (Call that your x-axis.) Also you have m2 in there for some reason.
So I figure a1 = 2/3(a2)?
Better rethink that.
After that I'm just getting lost in a mess of algebra...
The algebra will be much simpler when you change your x-axis for m1.
 
OK.

I advise you to consider components parallel to the incline surface. (Call that your x-axis.) Also you have m2 in there for some reason.

Better rethink that.

The algebra will be much simpler when you change your x-axis for m1.

so if it's parallel then the summation of forces on m1 just becomes the other way around with mg(cos theta) and mg(sin theta). How is that simpler?

If there's two forces of tension acting on m1 and three acting on m2 shouldn't that be the acceleration?
 

Doc Al

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so if it's parallel then the summation of forces on m1 just becomes the other way around with mg(cos theta) and mg(sin theta). How is that simpler?
It's simpler because the acceleration of m1 is parallel to the incline.

If there's two forces of tension acting on m1 and three acting on m2 shouldn't that be the acceleration?
No. What you're trying to figure out is the constraint due to the fact that the masses are connected by a rope. If m1 moves up the incline by a distance X, how far down does m2 move?
 
If m1 goes up the plane by 1 meter then m2 will go down by 1/3 of a meter?
 

Doc Al

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If m1 goes up the plane by 1 meter then m2 will go down by 1/3 of a meter?
No. Think of it this way. Consider the pulley at the top of the incline. Say a length of rope X moves to the right over that pulley. Considering how the rope to m1 is folded over, how far does m1 move up? Similarly, how far does m2 move down?
 
No. Think of it this way. Consider the pulley at the top of the incline. Say a length of rope X moves to the right over that pulley. Considering how the rope to m1 is folded over, how far does m1 move up? Similarly, how far does m2 move down?
m1 will move up 1/2X and m2 will move down 2/3X?
 

Doc Al

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m1 will move up 1/2X and m2 will move down 2/3X?
Half right. Since for m1 the length X is distributed over two rope segments, m1 will move up 1/2 X. But for m2 that same length is distributed over three rope segments, so m2 moves down 1/3 X.
 
Half right. Since for m1 the length X is distributed over two rope segments, m1 will move up 1/2 X. But for m2 that same length is distributed over three rope segments, so m2 moves down 1/3 X.
Ok. So now

m1a1 = 2T-m1g(sin(theta))
m2a2 = 3T-m2g

a1= 1.5(a2)?

T = (-m1g(sin(theta))-m1a1)/2
T = (-m2g-m2a2)/3

3(-m1g)(sin(theta))-3(m1a1) = 2(-m2g)-2(m2a2)

and substitute a1 for a2 and vice versa to solve?
 

Doc Al

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Ok. So now

m1a1 = 2T-m1g(sin(theta))
OK.
m2a2 = 3T-m2g
OK.

In both cases you took up as positive, which is fine. (Another way is to use up as positive for m1 and then down as positive for m2. That will let both a1 and a2 be positive.)

a1= 1.5(a2)?
That's the correct relationship for the magnitudes, but you also have to worry about the sign. Since you are using up as positive, you need to use a1= -1.5(a2) to capture the fact that when m1 moves up, m2 moves down.

You'll end up with three equations and three unknowns, which you can solve by substitution.
 
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Alright, I'm a bit confused when it comes to the algebra.

I got a2 = (2(m2)+3g(sinTheta)-2g)/6.5
 

Doc Al

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Alright, I'm a bit confused when it comes to the algebra.

I got a2 = (2(m2)+3g(sinTheta)-2g)/6.5
You must have made an error, since your first term has units of mass, not acceleration. And your result does not contain m1, which cannot be correct.
 
3(-m1*gsinθ)- 3(m1a1) = 2(-m2g)-2(m2a2)
3(-m1*gsinθ)- 3(m1*-1.5a2) = 2(-m2g)-2(m2a2)

I'm not sure what what do after this point.
 

Doc Al

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3(-m1*gsinθ)- 3(m1a1) = 2(-m2g)-2(m2a2)
3(-m1*gsinθ)- 3(m1*-1.5a2) = 2(-m2g)-2(m2a2)

I'm not sure what what do after this point.
Get rid of all those extra negative signs for one thing. You can rewrite that first line as:
3(m1*gsinθ) + 3(m1a1) = 2(m2g) + 2(m2a2)

Once you get your second equation, it just has one unknown--a2. To solve for a2, just move all the terms containing a2 to one side.
 
Get rid of all those extra negative signs for one thing. You can rewrite that first line as:
3(m1*gsinθ) + 3(m1a1) = 2(m2g) + 2(m2a2)

Once you get your second equation, it just has one unknown--a2. To solve for a2, just move all the terms containing a2 to one side.
Which 2nd equation :confused:
 

Doc Al

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Which 2nd equation :confused:
The one where you've eliminated a1.

In your last post you had two equations:
3(-m1*gsinθ)- 3(m1a1) = 2(-m2g)-2(m2a2)
That's the first.
3(-m1*gsinθ)- 3(m1*-1.5a2) = 2(-m2g)-2(m2a2)
That's the second. That's the one I meant.

Just get rid of the extraneous minus signs--they make things look harder than they are. You'd never write an equation like -a = -b when you can simply write a = b. In your 2nd equation you can get rid of 4 out of 5 minus signs.

Other than that you're doing fine.
 

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