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Very basic vector/force problem

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  1. Aug 17, 2015 #1
    I have a question about sign when it comes to solving vector problems. Here is one for example:
    A 5.0-kilogram mass is positioned on a 30 degree incline. It is kept stationary by a string pulling parallel to the incline. Determine the tension of the string.

    This is a very easy problem. I started by defining down the incline as positive and up the incline as negative. Thus,

    ##F_{net} = F_{g\parallel } - T##
    ##0 = F_{g\parallel } - T##
    ## T = F_{g\parallel } ##
    ##T = mg\sin\theta##
    ##T = 25 N##

    I am getting 25 newtons, not -25 newtons. Why is this? The force of the tension, which is in the opposite direction as the parallel component of gravity (+25 N), should be negative, since it is cancelling the positive force. Where am I going wrong in the calculation?
     
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  3. Aug 17, 2015 #2

    RUber

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    This is happening because you put a negative sign in front of T. If T and F are supposed to point in opposing directions, then 0 = F + T.
     
  4. Aug 17, 2015 #3
    So when I write the equation down, am I supposed to do ##F_{net} = F_{g\parallel } - T## or ##F_{net} = F_{g\parallel } + T##?
     
  5. Aug 17, 2015 #4

    RUber

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    It is the sum of the forces, F + T.
     
  6. Aug 17, 2015 #5

    jbriggs444

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    ##F_{net} = F_1 + F_2 + ...##

    If you use the letter T to denote the force resulting from tension then T will have a negative value. The equation you get will be ##F_{net} = F_1 + T##

    If you use the letter T to denote the tension in the string and if you want T to have a positive value then the force from tension, ##F_t## will be given by -T and the equation you get will be ##F_{net} = F_1 + F_t = F_1 - T##
     
  7. Aug 17, 2015 #6
    So whenever I am doing a dynamics problem, no matter the case, I should sum the forces? Even if they are pointing in opposite directions?
     
  8. Aug 17, 2015 #7

    RUber

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    The sign of the force should be included. So if gravity is pulling with 25N in the positive direction and T is pulling with 25N in the negative direction,
    F_g = 25N, T = -25N.
    F_g + T = 0.
    It is all dependent on how you define your forces, like in the example by jbriggs.
     
  9. Aug 17, 2015 #8

    A.T.

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    The net force is the vector sum of all forces. The vectors might have negative and positive components.
     
  10. Aug 17, 2015 #9
    Said another way, you can draw your forces any way you want, you can define your positive and negative direction anyway you want....for as long as you write your equations correctly according to your own assumptions, the sign of the result indicates whether the actual force is in the direction you assumed it was (results positive) or opposite to what you assumed (results negative).
     
  11. Aug 17, 2015 #10

    FactChecker

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    It is best to have one direction positive and use that for all forces. You chose down to be positive for F, so make down positive for T. Then T will end up negative if it points up. If you change directions and coordinate systems much, it can get unnecessarily confusing very fast.
     
  12. Aug 17, 2015 #11

    Simon Bridge

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    The confusion arises because you did not write down a vector equation right at the start.
    The method you used letters to represent the magnitudes of the vectors, while the sign recorded the direction separately. This is a good habit for making fewer algebraic errors.
    The full vector equation is the one with plus signs between the letters, which should also have some other notation to indicate they are vectors. In your example, there are three vectors that sum to zero... not two.
     
  13. Aug 17, 2015 #12

    jbriggs444

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    On a somewhat different matter...

    I cringe a bit when the letter T is used as the name of a force and is given a direction. The letter T suggests "tension". But the tension in a string is always positive and is directionless. Strings only support tension, not compression. The tension in a string does not point in either direction specifically. Instead, the direction of the associated force comes from the direction of the attachment point. The force from tension exerted by a string on the object to which it is attached is always exerted in the direction from the attachment point along the axis of the string. e.g. one team in a tug-of-war is pulled east by the rope and the other team is pulled west but the tension in the rope is neither east nor west specifically.

    Back to the case at hand. Once you have determined the force from tension and found (using the downward = positive convention) that it is negative you could note that the direction from attachment point along the string is also negative. Thus a positive tension gives rise to a negative force from tension. i.e. ##F_t## is negative but ##T## is positive.
     
  14. Aug 25, 2015 #13
    As Simon Bridge pointed out, you should start with a vector equation. Make a clear distinction between vectors and components. Newton's law says you ADD the force VECTORS to get mass times the acceleration VECTOR. When you write this down, realize that there is nothing called a positive vector or a negative vector, whichever way it is pointing. Vectors are not ordered. A vector pointing downward is not negative, and one pointing upward is not positive.

    The next step is to take components of your vector equation along specified coordinate axes. A vector may have a positive or a negative component depending on the direction of the axis relative to that vector. For example, if you take the y axis pointing downward, then the y component of your weight is a positive number.

    A discussion of whether the tension in a string is positive or negative is irrelevant. The question that should be asked is, "what is the direction of the force (called tension) and is its component along a particular axis positive or negative?"
     
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