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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
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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perl
 one year ago
Best ResponseYou've already chosen the best response.1in both cases Tension points up m1*g points down m2*g points down

perl
 one year ago
Best ResponseYou've already chosen the best response.1@Michele_Laino @IrishBoy123

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4In 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.

perl
 one year ago
Best ResponseYou've already chosen the best response.1hmm, i thought you can't change the direction of a force arbitrarily .

perl
 one year ago
Best ResponseYou've already chosen the best response.1i mean once you set one coordinate frame

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4They're two different equations. And they've set two different signconventions for each of them. As long as you're following the convention within an equation, it is legal.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4You can set the same signconvention for both equations, but you'd still get the same thing.

perl
 one year ago
Best ResponseYou've already chosen the best response.1if you set the same sign convention for both equations, it would be different

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4It would remain the same.

perl
 one year ago
Best ResponseYou've already chosen the best response.1Fnet = T  M1g Fnet = T  m2g

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4In one equation, the acceleration is positive, while in the other, it is negative.

perl
 one year ago
Best ResponseYou've already chosen the best response.1ok if i changed the problem a bit, using a horizontal instead of two vertical, you would use the same logic. dw:1433096368815:dw

perl
 one year ago
Best ResponseYou've already chosen the best response.1that does seem true that in the first case acceleration is up, and in the second case acceleration is down

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4Yes, and you get the same system of equations in the end.

perl
 one year ago
Best ResponseYou've already chosen the best response.1but... the sign of the acceleration depends on the net force

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4dw:1433096531777:dw

perl
 one year ago
Best ResponseYou've already chosen the best response.1the sign of the acceleration is in the same direction as the net force

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4Alright, I will explain that.

perl
 one year ago
Best ResponseYou've already chosen the best response.1ok 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?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.4In 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.