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

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- perl

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

- katieb

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- perl

in both cases Tension points up
m1*g points down
m2*g points down

- 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

hmm, i thought you can't change the direction of a force arbitrarily .

- perl

i mean once you set one coordinate frame

- 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

You can set the same sign-convention for both equations, but you'd still get the same thing.

- perl

if you set the same sign convention for both equations, it would be different

- ParthKohli

It would remain the same.

- perl

Fnet = T - M1g
Fnet = T - m2g

- ParthKohli

In one equation, the acceleration is positive, while in the other, it is negative.

- 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

that does seem true that in the first case acceleration is up, and in the second case acceleration is down

- ParthKohli

Yes, and you get the same system of equations in the end.

- perl

but... the sign of the acceleration depends on the net force

- ParthKohli

|dw:1433096531777:dw|

- perl

the sign of the acceleration is in the same direction as the net force

- ParthKohli

Alright, I will explain that.

- 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

|dw:1433096693101:dw|

- 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

|dw:1433096785244:dw|