Atwood machine tension as m1 approaches infinity

[Meteo]

An atwood machine holds two masses at its ends by a massless rope. m1 and m2. Assume the pully is frictionless.
What will the magnitude of the tension(T) go to as m1 goes to infinity but m2 stays the same?
Im very lost on this problem. I have drawn a free body diagram showing m1g and m2g pointing downwards and T pointing upwards.

I have derived two equations

$$m_1g-T=-m_1a$$ where m1 is going down. I subtracted T from m1g because m1g is much bigger.
$$T-m_2g=m_2a$$ where m2 is going up. I did this because T should be greater than m2g. i think.
but apparently that's not right, and I don't know why.

 

[Tide]

You need to analyze your free body diagram a little more carefully!

Alternatively, you should realize that in the limit of large $m_1$, $m_2$ will be accelerated upward at a rate g, i.e. it pulls on the string with a force equal to $m_2 g$ due to the acceleration. That, plus its weight, gives you the tension.

 

[Meteo]

why is m2 accelerated upwards at g?
its weight will be m2g... so $$T=2m_2g$$?

I think I am comfused on what tension is...so the force of it pulling up and the force of the weight pulling down = tension?

 

[Diane_]

Tide was talking about the limiting case as m1 gets very large. Suppose m1 were 30,000 tons and m2 were 2 ounces. Can you see that m2 would make very little difference in that case? So, for all practical purposes, m2 would accelerate upwards at g because m1 would accelerate downwards at g.

As for what tension is - well, you can consider it a force exerted by the rope. It'll pull both m1 and m2 upwards - exactly how much depends on how m1 and m2 pull downward on their ends.

Go back and do as Tide suggested - analyze the free body diagram. But when you do it, try to picture it as a physical system rather than just a system of equations. If the math confuses you, the physics will always pull you out. Just ask yourself what should happen.

 

[Tide]

$m_2$ is accelerated upward at g because it offers negligible resistance to $m_1$ falling.

The tension is the force exerted by the string on the mass. The forces acting on the mass are the tension and force of gravity so $T - m_2 g = m_2 a$. But a = g, therefore ...!

 

[Tide]

Meteo,

Listen to Diane - she's good! :)

 

[Diane_]

How does one blush online? :)

 

[Tide]

Like this?

I didn't mean to make you blush - just offering some sound advice!