Acceleration of 10 kg mass on frictionless surface with pulley and 5 kg hanging mass

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grouchy
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For the system in the figure mass m1 is 10 kg and mass m2 is 5 kg. What is acceleration of m1 if the horizontal surface is frictionless? Hint: How does a1 compare to a2?


http://geocities.com/grouchy187/untitled.bmp

My attemp:
T = m1a T - m2g = m2a
m1a = m2g + m2a
a= m2g / m1-m2
 

Discussion

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What tension force acts on m2? (Consider the pulley as part of m2.)

What about that hint?
 
for m2, would the weight W, be equal to 2T?

also I think a1 = a2
 
grouchy said:
for m2, would the weight W, be equal to 2T?
No. But those are the forces acting on m2. (They would only be equal if m2 were not accelerating.)

also I think a1 = a2
No. Try to think this through. (Using a piece of string to work it out may help--I'm not kidding.) If m1 moves 1 m to the right, how far down does m2 move?
 
i dunno, I'm thinking about it and I just don't get this one
 
If m2 moves down by 1 meter, how much extra rope will hang down off the table? That extra amount of rope is the distance that m2 moves (since m2 is tied directly to the rope). Once you have that, the ratio of distances in this case is the same as the ratio of accelerations.

If you're having trouble visualizing the amount of extra rope, just start with your orginal diagram. Now draw in the new position of m2 and the lower pulley 1 meter below where they were. Note that you have to extend the lines for the ropes on each side of the pulley. How much extra rope do these new lines represent?