- #1

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Is this right?

Thanks,

Dorothy

- Thread starter Dorothy Weglend
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- #1

- 247

- 2

Is this right?

Thanks,

Dorothy

- #2

Doc Al

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The blocks and pulley are connected, so the acceleration of all three must be aligned.

- #3

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Thanks, Doc Al. Well, here is my problem, then.

The blocks and pulley are connected, so the acceleration of all three must be aligned.

A m1 is connected over a pulley to a m2 on an incline. Let's say the incline is to the right of the pulley, so that the acceleration is to the right.

T1 - friction1 = m1a

m2g(sin theta) - friction2 - T2 = m2a

(m1 + m2)R = -I(alpha) = -Ma/2 (where M is mass of pulley).

If I rotate the diagram 180 degrees, so the incline is on the left, acceleration is to the left, keeping positive in the direction of the acceleration, the only equation that would change is the torque:

(m1 + m2)R = Ma/2

Which is obviously going to lead to a different acceleration by just changing the horizontal orientation, which is a physical impossibility, of course.

Or I'm confused (probably)... Clear this up for me, please?

Thanks, as always,

Dorothy

- #4

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Oh, I'm thinking m1 is on a horizontal surface, if that makes any difference.

- #5

Doc Al

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These two make sense.T1 - friction1 = m1a

m2g(sin theta) - friction2 - T2 = m2a

But this one doesn't. Why are m1 and m2 mentioned in the equation for the pulley?(m1 + m2)R = -I(alpha) = -Ma/2 (where M is mass of pulley).

What forces act on the pulley? What torques do they exert? Set the net torque equal to I(alpha).

This is an artifact of this strange equation. Fix it.If I rotate the diagram 180 degrees, so the incline is on the left, acceleration is to the left, keeping positive in the direction of the acceleration, the only equation that would change is the torque:

- #6

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I'm an idiot. I meant to write:These two make sense.

But this one doesn't. Why are m1 and m2 mentioned in the equation for the pulley?

What forces act on the pulley? What torques do they exert? Set the net torque equal to I(alpha).

This is an artifact of this strange equation. Fix it.

(T2 - T1)R = Ma/2

And really, that's what I was thinking. My fingers typed something else. :yuck:

But that still leaves me with the same question, doesn't it?

With the incline on the right, (T2 - T1)R = -Ma/2

With it on the left, (T2 - T1)R = Ma/2

Since the only that has changed is the sense of the torque?

- #7

Doc Al

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OK.With the incline on the right, (T2 - T1)R = -Ma/2

Nope. The signs of T2 and T1 will change also.With it on the left, (T2 - T1)R = Ma/2

If the sense of the torques change, so must their signs.Since the only that has changed is the sense of the torque?

- #8

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But how is that possible? Don't the signs of T2 and T1 (the tensions on the cord) depend on the sign of the acceleration? T2 will always be positive, since it is pulling on the pulley in the direction of a, and T1 will always be negative.OK.

Nope. The signs of T2 and T1 will change also.

If the sense of the torques change, so must their signs.

Gosh, I'm sorry to be difficult, but I am just not getting this.

- #9

Doc Al

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But why did you change the sign of the acceleration but not the torques?But how is that possible? Don't the signs of T2 and T1 (the tensions on the cord) depend on the sign of the acceleration? T2 will always be positive, since it is pulling on the pulley in the direction of a, and T1 will always be negative.

If all you did was flip the diagram, so that the acceleration now goes left instead of right, then you can:

(1) Use a convention that right is + and left is -, in which case the signs of T1, T2, and "a" all get reversed;

(2) Stick to the convention that "a" is always positive, in which case all the signs stay the same.

Pick one!

Either way, it's all or nothing.

- #10

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Well, I picked (2). But the sign of the torque has to change, right? If the acceleration is to the right, won't the pulley be rotating clockwise?But why did you change the sign of the acceleration but not the torques?

If all you did was flip the diagram, so that the acceleration now goes left instead of right, then you can:

(1) Use a convention that right is + and left is -, in which case the signs of T1, T2, and "a" all get reversed;

(2) Stick to the convention that "a" is always positive, in which case all the signs stay the same.

Pick one!

Either way, it's all or nothing.

Now, if I flip the diagram, keeping convention 2, then the pulley is rotating counterclockwise. Doesn't that change the sign of the torque?

Which is why I thought I should ignore the sense of the torque in a problem like this, and always treat it as a positive number. Ok, I should give it the same sign as the acceleration, in this case, positive. Well, in spite of your good efforts, I am still thinking this

- #11

Doc Al

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The sign of the torque depends on your convention. Using convention 2, if it accelerates clockwise then clockwise is positive. Everything flips, so the signs all stay the same.Well, I picked (2). But the sign of the torque has to change, right? If the acceleration is to the right, won't the pulley be rotating clockwise?

Now, if I flip the diagram, keeping convention 2, then the pulley is rotating counterclockwise. Doesn't that change the sign of the torque?

Counting clockwise as + and counterclockwise as - (or vice versa) is the same as using convention 1, not convention 2.

Giving torque the same sign as the acceleration is not the same as "ignoring the sense of the torque"--it's following convention 2.Which is why I thought I should ignore the sense of the torque in a problem like this, and always treat it as a positive number. Ok, I should give it the same sign as the acceleration, in this case, positive. Well, in spite of your good efforts, I am still thinking this

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- #12

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Thank you for your patience and help, as always.

Dorothy

- #13

Doc Al

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You are *always *welcome, Dorothy!

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