Acceleration along a pulley with a ledge

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fontseeker

Homework Statement



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Homework Equations



F = ma

The Attempt at a Solution



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The solution to the problem is 1.12m/s^2. However, I don't know how they got to that number. Every way I do it I end up with 3.2m/s^2.
 

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Doc Al said:
Hint: Do the masses have the same acceleration?
They do, so that is why I am treating acceleration with the same variable for both. The answer book says they have the same acceleration but in different direction, but the magnitude should be the same.
 
fontseeker said:
They do, so that is why I am treating acceleration with the same variable for both. The answer book says they have the same acceleration but in different direction, but the magnitude should be the same.
Rethink that. If the 20 kg mass moves 1 m to the right, how far does the other mass move down? (Consider the string.)
 
Doc Al said:
Rethink that. If the 20 kg mass moves 1 m to the right, how far does the other mass move down? (Consider the string.)
0.5m down. However, that means that they are both positive then? However, I don't know how to prove it, it just makes sense since the tension is twice as much on the 20kg object.
 
fontseeker said:
0.5m down.
No. If the 20 kg mass moves 1 m to the right, how much rope must have passed over the pulley?
 
Doc Al said:
No. If the 20 kg mass moves 1 m to the right, how much rope must have passed over the pulley?
One meter would have had to pass.
 
fontseeker said:
One meter would have had to pass.
Nope. Draw yourself a quick picture of before and after and measure the length of the string that must have passed. (Or grab an actual piece of string and try it.)
 
Doc Al said:
Nope. Draw yourself a quick picture of before and after and measure the length of the string that must have passed. (Or grab an actual piece of string and try it.)
Ah okay, since it is moving to the right 1 meter, 1 meter at the top and 1 meter at the bottom are loose, meaning 2 meters down.
 
fontseeker said:
Ah okay, since it is moving to the right 1 meter, 1 meter at the top and 1 meter at the bottom are loose, meaning 2 meters down.
Good! So how must the accelerations of the two masses relate to each other?
 
Doc Al said:
Good! So how must the accelerations of the two masses relate to each other?
The acceleration of the second block would be twice the acceleration of the first.
 
fontseeker said:
The acceleration of the second block would be twice the acceleration of the first.
Exactly. Now rewrite your equations with that in mind.
 
Doc Al said:
Exactly. Now rewrite your equations with that in mind.
I get 2.45m/s^2, which is the answer I see all over the internet. However, on my book this is what it says:

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fontseeker said:
I get 2.45m/s^2, which is the answer I see all over the internet.
That's the correct answer for the acceleration of the 20 kg mass.

fontseeker said:
However, on my book this is what it says:
That answer makes no sense for this problem. What book is this?
 
Doc Al said:
That's the correct answer for the acceleration of the 20 kg mass. That answer makes no sense for this problem. What book is this?
Tipler
 
fontseeker said:
Tipler
I have the 6th edition, which has that as problem 4-81. The answers seem OK to me.