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Solving for Tension in Net Force Equation: Need Help!
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[QUOTE="kuruman, post: 6850717, member: 192687"] I agree with [USER=334404]@haruspex[/USER] that your equation is ambiguous but not because it depends on whether you define the acceleration positive when it is up or positive when it is down. It's because, apparently, you didn't incorporate the known direction of the acceleration in your equation. See the explanation below which is lengthier than originally planned and I apologize for that. Here we have a 1 D vector equation expressing Newton's second law. Using arrows over symbols to indicate vectors, $$\sum_i \vec{F}_i=m\vec a.$$Now $$\vec T+m\vec g=m\vec a$$Note that a "plus" sign appears between all vectors because the net force is the sum of all forces in Newton's law. In 1D, the equation is simplified by removing the arrows to convert the vector equation into a scalar equation. There are two possibilities. [B]Case I[/B]: You know the directions of all vectors in the 1D equation. First you pick a direction as positive. Second, you replace the arrows with "+" sign in front of the vector symbol if it points along the chosen positive direction and a "-" sign if the vector points opposite to the chosen positive direction. In this particular problem you know that the direction of the tension is "up", the direction of the weight is "down" and the direction of the acceleration is also "down." Following this rule, [LIST=1] [*]If you choose "up" as positive ##~\vec T+m\vec g=m\vec a \rightarrow T-mg=-ma.## [*]If you choose "down" as positive ##~\vec T+m\vec g=m\vec a \rightarrow -T+mg=ma.## [/LIST] Note that one equation becomes the other if you multiply both sides by ##-1## which is equivalent to choosing the opposite direction as positive. Also, symbols ##T##, ##g## and ##a## in the equations stand for magnitudes of vectors and are positive quantities. If you replace symbols with given numbers in either equation and the acceleration comes out negative, you will know that you made a mistake somewhere. The magnitude of a vector is positive independently of the choice of axes. [B]Case II[/B]: You don't know the direction of one vector in the 1D equation. You follow the same rule with the additional directive "keep it simple". We can modify the situation in this problem a bit and say the we have a vertically accelerating mass ##m## attached to a string in which the tension is ##T##. We are asked to find the acceleration, a vector, in terms of the other quantities. First note that the tension is "up" and the weight are "down", therefore the tension and the weight must appear with a relative negative sign in the expression for the net force regardless of whether "up" is positive or negative. Newton's second law can be written in vector equation form (a bit unconventionally but clearly as to what is meant) $$\frac{1}{m}T\text{(up)} +g\text{(down)}=a\text{(?)}$$ We can remove the directions that we know and replace them with negative signs. For the acceleration, we "keep it simple" and assume that the unknown direction denoted by the question mark is positive as defined by the choice of positive. Note that here ##a## stands for the scalar component of the vector not its magnitude; it could be positive or it could be negative, we just don't know. We get [LIST=1] [*]##a=\dfrac{1}{m}T -g~## if "up" is positive [*]##a=-\dfrac{1}{m}T +g~## if "down" is positive [/LIST] Either equation is correct and equation 1 is the form that you have posted. Note that one does not get one equation from the other if both sides are multiplied by ##-1##, i.e. they are not the same equation. That is, I think, what [USER=334404]@haruspex[/USER] meant when he said Nevertheless, if you substitute numbers in either equation and the acceleration turns out to be a negative number, it will mean that the initial assumption (when the question mark was removed) was incorrect. The acceleration must be opposite to the chosen positive direction. The catch here is that if you make a calculation error and the acceleration comes out with one sign when it should be the other, you wouldn't know it. That is why if you know the direction of the acceleration, you should put it in as shown in case I. [/QUOTE]
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Solving for Tension in Net Force Equation: Need Help!
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