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perl

  • one year ago

Pulley question, why does sign of Tension and m*g change in the diagram http://prntscr.com/7bk4a2

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  1. perl
    • one year ago
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    in both cases Tension points up m1*g points down m2*g points down

  2. perl
    • one year ago
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    @Michele_Laino @IrishBoy123

  3. ParthKohli
    • one year ago
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    In one equation, they've considered upwards as positive and in the other one, they've considered upwards as negative. In either case, they've selected the direction of acceleration as the positive one.

  4. perl
    • one year ago
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    hmm, i thought you can't change the direction of a force arbitrarily .

  5. perl
    • one year ago
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    i mean once you set one coordinate frame

  6. ParthKohli
    • one year ago
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    They're two different equations. And they've set two different sign-conventions for each of them. As long as you're following the convention within an equation, it is legal.

  7. ParthKohli
    • one year ago
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    You can set the same sign-convention for both equations, but you'd still get the same thing.

  8. perl
    • one year ago
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    if you set the same sign convention for both equations, it would be different

  9. ParthKohli
    • one year ago
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    It would remain the same.

  10. perl
    • one year ago
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    Fnet = T - M1g Fnet = T - m2g

  11. ParthKohli
    • one year ago
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    In one equation, the acceleration is positive, while in the other, it is negative.

  12. perl
    • one year ago
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    ok if i changed the problem a bit, using a horizontal instead of two vertical, you would use the same logic. |dw:1433096368815:dw|

  13. perl
    • one year ago
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    that does seem true that in the first case acceleration is up, and in the second case acceleration is down

  14. ParthKohli
    • one year ago
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    Yes, and you get the same system of equations in the end.

  15. perl
    • one year ago
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    but... the sign of the acceleration depends on the net force

  16. ParthKohli
    • one year ago
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    |dw:1433096531777:dw|

  17. perl
    • one year ago
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    the sign of the acceleration is in the same direction as the net force

  18. ParthKohli
    • one year ago
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    Alright, I will explain that.

  19. perl
    • one year ago
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    ok i guess that makes sense, since in diagram 1 the first net force on the left is up and the second net force is down?

  20. perl
    • one year ago
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    |dw:1433096693101:dw|

  21. ParthKohli
    • one year ago
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    In your case (the screenshot), the first equation has considered up to be + and down to be -. \[F_{net}=T - m_1 g = m_1 (+a)\]Here, acceleration is upwards. Now the second one has considered up to be - and down to be +. The equation then becomes\[F_{net} = -T + m_2 g = m_2(+a)\]The main aim was to get acceleration to be positive. If we follow the convention of the previous case, then the equation is\[T - m_2 g = m_2(-a)\]Which doesn't make any difference.

  22. perl
    • one year ago
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    |dw:1433096785244:dw|