perl
  • perl
Pulley question, why does sign of Tension and m*g change in the diagram http://prntscr.com/7bk4a2
Physics
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perl
  • perl
Pulley question, why does sign of Tension and m*g change in the diagram http://prntscr.com/7bk4a2
Physics
katieb
  • katieb
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perl
  • perl
in both cases Tension points up m1*g points down m2*g points down
perl
  • perl
ParthKohli
  • ParthKohli
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.

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perl
  • perl
hmm, i thought you can't change the direction of a force arbitrarily .
perl
  • perl
i mean once you set one coordinate frame
ParthKohli
  • ParthKohli
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.
ParthKohli
  • ParthKohli
You can set the same sign-convention for both equations, but you'd still get the same thing.
perl
  • perl
if you set the same sign convention for both equations, it would be different
ParthKohli
  • ParthKohli
It would remain the same.
perl
  • perl
Fnet = T - M1g Fnet = T - m2g
ParthKohli
  • ParthKohli
In one equation, the acceleration is positive, while in the other, it is negative.
perl
  • perl
ok if i changed the problem a bit, using a horizontal instead of two vertical, you would use the same logic. |dw:1433096368815:dw|
perl
  • perl
that does seem true that in the first case acceleration is up, and in the second case acceleration is down
ParthKohli
  • ParthKohli
Yes, and you get the same system of equations in the end.
perl
  • perl
but... the sign of the acceleration depends on the net force
ParthKohli
  • ParthKohli
|dw:1433096531777:dw|
perl
  • perl
the sign of the acceleration is in the same direction as the net force
ParthKohli
  • ParthKohli
Alright, I will explain that.
perl
  • perl
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?
perl
  • perl
|dw:1433096693101:dw|
ParthKohli
  • ParthKohli
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.
perl
  • perl
|dw:1433096785244:dw|