Double pulley problem-difference in acceleration

[Arait]

Double pulley problem--difference in acceleration

Homework Statement

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Homework Equations

a= g(M-m)/M+m

The Attempt at a Solution

Okay, so overall, pulleys just confuse me. We barely mentioned them in class and there's precious little about them in our books. I tried to answer this question and was able to get the first part (a) by finding the above equation on Google. However, as the question would lead you to assume, the answer to be is NOT the same. The problem is, I see no difference other than the fact that a box was replaced by hand. I don't understand why, if the downward force is the same, you would get a different acceleration. Because I don't understand this, it's making it pretty much impossible to know what to change in order to solve part b. I don't even know where to get started...

 

[Doc Al]

a= g(M-m)/M+m

You should learn how this formula was derived using Newton's 2nd law.

To see the difference between the two situations, compare the tension in the rope. Are they the same? Is the tension in version a equal to the weight of the second mass?

 

[Arait]

I'm still lost. The mass of the hand is probably different from the mass of the block, but as long as the force is the same, I don't see how it would change anything.

 

[Arait]

I found something in my book that says that a the force T applied at one end of amassless rope is transmitted undiminished to the other end. Does this mean that T=958 N on both sides?

 

[Doc Al]

I'm still lost. The mass of the hand is probably different from the mass of the block, but as long as the force is the same, I don't see how it would change anything.

True, if the force were the same then there would be no difference in the two scenarios. But is the force the same?

Answer my question: Is the tension in version a equal to the weight of the second mass?

 

[Doc Al]

I found something in my book that says that a the force T applied at one end of amassless rope is transmitted undiminished to the other end.

Yes, for a massless rope the tension is the same throughout.

Does this mean that T=958 N on both sides?

What makes you think that that's the tension?

 

[Arait]

Yes, for a massless rope the tension is the same throughout.

What makes you think that that's the tension?

I don't know. It's the opposite of gravity? I know that doesn't work since there's acceleration other than gravity acting in the downward direction, but I don't know how to solve for tension when there are so many variables you don't know.

 

[Arait]

True, if the force were the same then there would be no difference in the two scenarios. But is the force the same?

Answer my question: Is the tension in version a equal to the weight of the second mass?

I'm betting it's not... But I don't know how to calculate it.

 

[Doc Al]

I don't know. It's the opposite of gravity? I know that doesn't work since there's acceleration other than gravity acting in the downward direction, but I don't know how to solve for tension when there are so many variables you don't know.

You can solve for the tension and the acceleration--which are the only two unknowns in case a--by applying Newton's 2nd law to each mass. Start by identifying the forces on each.

(In case b you are given the tension, so it should be easier to calculate the acceleration.)

 

[Arait]

I got it right by taking 958-(small mass)(g)=(small mass)(a). I got a=12.99 m/s^2. So, my homework is now done, but even though I figured out the right number, I'm still conceptually having a problem seeing the difference between Case A and Case B.

 

[Doc Al]

I got it right by taking 958-(small mass)(g)=(small mass)(a). I got a=12.99 m/s^2.

That's not quite correct. Is it reasonable that the acceleration of the masses would be greater that something in free fall?

Show how you arrived at this result.

So, my homework is now done, but even though I figured out the right number, I'm still conceptually having a problem seeing the difference between Case A and Case B.

You'll need to redo your solution for case a. Then you can compare the tension in each case. (What's the tension in case b?)

 

[Arait]

I think I kinda see it now. Thanks!