Calculating Mass of Atwood Machine to Keep It Still

[mmmboh]

In the Atwood machine, what should M be, in terms of m1 and m2 so that it doesn't move? [PLAIN]http://img163.imageshack.us/img163/8288/atwood.jpg

My work: It doesn't move, so I said T₁=Mg (T₁ is the tension in the rope attached to M), I believe the tension in the rope connecting m1 and m2 is the same since its the same rope, so I said T₁=2T₂. And I said T₂-m1g=m1*a, and T₂-m2g=-m2*a, negative because the acceleration of those masses is in opposite directions, but should be of the same magnitude. Adding those two equations together, I find T₂=[m1(g+a)+m2(g-a)]/2, and since T₁=2T₂=Mg, I said
M=[m1(g+a)+m2(g-a)]/g...and I can simplify this to M=2m₂*(g-a)/g.
I don't believe this is right because it isn't just in terms of m1 and m2...I did it another way before and got M=(m1+m2)/2, but I don't believe the process was right.
Any help? this is due in the morning.

Thanks.

 

[PhanthomJay]

In the Atwood machine, what should M be, in terms of m1 and m2 so that it doesn't move?

My work: It doesn't move, so I said T₁=Mg (T₁ is the tension in the rope attached to M), I believe the tension in the rope connecting m1 and m2 is the same since its the same rope, so I said T₁=2T₂. And I said T₂-m1g=m1*a, and T₂-m2g=-m2*a, negative because the acceleration of those masses is in opposite directions, but should be of the same magnitude. Adding those two equations together, I find T₂=[m1(g+a)+m2(g-a)]/2, and since T₁=2T₂=Mg, I said
M=[m1(g+a)+m2(g-a)]/g...and I can simplify this to M=2m₂*(g-a)/g.
I don't believe this is right because it isn't just in terms of m1 and m2...I did it another way before and got M=(m1+m2)/2, but I don't believe the process was right.
Any help? this is due in the morning.

Thanks.

This is very good, and correct...but, you have written M as a function of m2, g, and a...you need to solve for a to find M as a function of m1, m2, and g
Edit...g cancels out...M is a function of m1 and m2 only

 

[mmmboh]

Ok, if I did my algebra right, I get 4m₁m₂/(m₁+m₂), is this what you got?

 

[PhanthomJay]

Ok, if I did my algebra right, I get 4m₁m₂/(m₁+m₂), is this what you got?

Yes, that is very good work, I must say. It is difficult setting up the equations, then doing the algebra, with letter vs. numerical values..nice job simplifying, also!

 

[rwisz]

I would first like to commend you for your efforts in attempting to solve this problem. Your approach is certainly on the right track, but there are a few issues with your calculations that may have led to incorrect results.

Firstly, in the Atwood machine, the masses m1 and m2 are connected by a rope that passes over a pulley. This means that the tension in the rope is the same throughout, as you correctly noted. However, your calculation of T1=2T2 is incorrect. The correct relationship is T1=T2+m1g, where T2 is the tension in the rope connected to m2. This is because the weight of m1 is pulling down on the rope connected to T1, increasing the tension in that rope by an amount equal to m1g.

Secondly, your equations for the tension in the two ropes are incorrect. The correct equations are T1=m1a and T2=m2a. This is because the acceleration of the masses is equal to the acceleration of the pulley, which is the same for both ropes. With these corrections, we can rewrite your equation for T2 as T2=[m1(g+a)-m2a]/2. Substituting this into T1=T2+m1g, we get T1=[m1(g+a)-m2a]/2+m1g.

Finally, to solve for M, we need to consider the forces acting on the entire system. The only external force acting on the system is the weight of M, which is Mg. Therefore, in order for the system to remain still, the net force on M must be zero. This means that T1-T2=Mg. Substituting our expressions for T1 and T2, we get:

[m1(g+a)-m2a]/2+m1g-[m1(g+a)-m2a]/2=Mg

Simplifying this equation gives us:

M=m1+m2

This result is in agreement with your previous calculation of M=(m1+m2)/2. This is because in a balanced Atwood machine, the mass of the pulley (represented by M) is equal to the average of the two masses on either side.

In conclusion, the correct expression for M in terms of m1 and m2 is M=m1+m2. I hope this clarifies any confusion and helps you to better understand the Atwood machine