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Doubt in resolving forces of a man in an elevator(normal force and weight)

  1. Jul 10, 2012 #1
    (Before I start, due to the use of vectors let me state I consider up to be positive and down to be negative)
    Imagine a scenario, of a man in an elevator. And the elevator is accelerating upwards. Now in this case the Normal force > weight. So Normal Force - Weight =mass X acceleration. If you think this is right, why? Shouldn't it become Normal Force + Weight = mass X acceleration, because for weight(mg) the g will be -9.8 and hence -m X -9.8, will give you a positive value of
    weight. I know its prolly wrong, because it would be insane to add Force and weight when their difference is what gives me the acceleration upwards, but then again why exactly is -9.8 wrong?
     
  2. jcsd
  3. Jul 10, 2012 #2
    Mass is scalar.
    How mass becomes -m?
     
  4. Jul 10, 2012 #3

    Doc Al

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    The best way to see it is as a vector equation:
    ƩF = Normal Force + Weight = ma
    Where 'normal force' and 'weight' are vectors.

    Let the magnitude of the normal force be called N; the magnitude of the weight is mg, where g = 9.8 m/s^2. (FYI: g does not have a minus sign!) Now using your sign convention, we can rewrite that vector equation as:
    ƩF = N -mg = ma
    or N = mg + ma
     
  5. Jul 10, 2012 #4
    it is -(m x -g) which will become +mg
     
  6. Jul 10, 2012 #5

    (FYI: g does not have a minus sign!)
    Why not?? It should, considering gravity is acting downwards!!
     
  7. Jul 10, 2012 #6

    A.T.

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    The constant g is a magnitude, not a vector.
     
  8. Jul 10, 2012 #7

    Doc Al

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    As A.T. explained, g is just the magnitude of the acceleration due to gravity--it does not have a sign. The full acceleration due to gravity with its appropriate sign would be -g. (Using up as positive.)

    The weight of the man is a downward force of magnitude mg. So using your sign convention you'd express that as -mg.
     
  9. Jul 10, 2012 #8
    Hmm..Usually in these situations when many people try to help, simple things get confusing.
    The OP chose the convention that up is positive. The forces are: normal-N, weight-W.
    Normal force acts on the body up, so N is positive. Weight acts downwards, so W is negative. And as the mass is always positive scalar, g is negative.
    So W=-9.81m.
    So I think, as Doc Al already pointed out, the best way to do similar problems is to first write down the force law vectorialy. And there there is a sum of vectors.
    Than as a second step you resolve the signs
     
  10. Jul 10, 2012 #9
    You are saying W=(-9.81m). Right? So now If i plug it into the equatio of N-W=ma won't I end up with N+W=ma ?? And maybe I know this already, but what exactly is the force law?
     
  11. Jul 10, 2012 #10
    he full acceleration due to gravity with its appropriate sign would be -g. So why are we not using the this? If it were then we would get (-W), and if we plug this value into the equation
    N-W= ma we would get N-(-W)=ma which would become N+W=ma. My method is to first find why the elevator is accelerating and hence get an appropriate equation(which I have done:N-W= ma). My next step is to plug in those values(with their vectorial signs), here is where I am stuck, as it become N + W.
     
  12. Jul 10, 2012 #11

    cepheid

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    Start with Newton's second law of motion: the sum of vertical forces is equal to mass times the acceleration. Note that this is a vector sum, which means that for a 1D problem, you have to take the sign of the force (up being positive and down being negative) into account.

    One of the points that others have been trying to make to you in this thread is that most people define the constant g as g = +9.81 m/s2, in which case the acceleration due to gravity is -g, and the weight is -mg. Keeping this in mind:

    ƩF = ma

    Fnorm + Fgrav = ma

    N - mg = ma

    N = ma + mg

    The *magnitude* of the normal force is equal to the *magnitude* of the weight plus the magnitude of the net force.

    EDIT: If you want to use the notation "W" for Fgrav, and N for Fnorm, that's fine. I myself switched to N halfway through to save typing.

    EDIT: I typed this before seeing that Doc Al typed nearly the exact same thing.
     
  13. Jul 10, 2012 #12
    F1+F2=ma is the standard equation if we do not know the direction of the vector.
    You shouldn't start with N-W=ma
    If W is in direction of N, which means added force, your equation still subracting.
    It means your equation N-W=ma to be standard form is wrong.
    Substituting N and F2=-W for the case on acceleration upward.
    N+(-W)=ma
     
    Last edited: Jul 10, 2012
  14. Jul 10, 2012 #13
    How about you just use forces as vectors. The net force, ma, is up because he is acclerating up. This is a positive number. Now what were the actual forces involved in accelerating the man up. Well, the earth is always pulling down on him with a force of mg. So mg is down. The normal force from the elevator floor on the man is up, so N is up positive. If you have one vector up, N, and another vector down, mg, what is the resultant vector?

    Its ma which is up. If you add the forces acting on the man you get N + (-mg) = ma If he was accelerating down you would get N + (-mg) = -ma. If he is not accelerating you get N + (-mg) = 0 oh and mg = W

    You are adding the Force vectors that are pointing in opposing directions on the man and getting a resultant that also has magnitude and direction.

    If you write N + W you are basically stating that both the normal force and weight are acting up? thats not what is happening.

    You might be best served by drawing a free body diagram and just work with forces as vectors.

    I apologize to others that have basically stated the same thing.
     
    Last edited: Jul 10, 2012
  15. Jul 10, 2012 #14
    infact g is a vector since its acceleration to gravity.
     
  16. Jul 11, 2012 #15

    A.T.

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    The constant g is the magnitude of that vector.

    That is how g is generally defined in those formulas. It doesn't make sense to have the direction, which depends on an arbitrary axis orientation choice, included in a constant. Because then it is not a constant anymore, but depends on your axis orientation.
     
    Last edited: Jul 11, 2012
  17. Jul 11, 2012 #16
    Basic mistake you are doing is you are considering the direction of weight as negative two times. You are taking weight as negative in equation(N-W=ma) because the acceleration(g) in weight formulae(mg) is negative. When you take g already negative in representing weight how can you take again g as negative. Remember weight and g are same except that they change in magnitude due to multiplication of mass m, which have magnitude but no direction.
     
  18. Jul 11, 2012 #17

    sophiecentaur

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    I realise we are all starting from different standpoints but there is no way that the direction of the Weight force (mg) is in doubt and, as m is definitely a scalar, then g must involve direction - i.e. it must be a vector. Also, g is also an acceleration, which, by definition, is a Vector.

    The only way an individual can sort this out in their minds is to draw a diagram of the forces involved and rigorously think in which direction they act. The equation of motion falls out of it. But you need to remember that the elevator is NOT in equilibrium if it is accelerating so you cannot write an equation based on that assumption.

    This is precisely what you need to do and you need to be strictly consistent, once you have chosen the reference frame. If you are, then your answer will be correct. You can confirm what I say by doing the sums using 'up=positive' and 'down=positive'. You will get the same answer if you are rigorous.
     
  19. Jul 11, 2012 #18

    Doc Al

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    The standard convention is to have g stand for the magnitude of the acceleration due to gravity. It has no direction; it's a constant, not a vector.

    The acceleration of a falling object is a vector with magnitude g and direction down.

    The weight of an object is a vector with magnitude mg and direction down.
     
    Last edited: Jul 11, 2012
  20. Jul 11, 2012 #19

    sophiecentaur

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    I can't disagree totally but somewhere along the line you need a direction / vector. If you are doing a ballistics problem, would you not put an arrow on the 'g' acting on the projectile? In the SUVAT approach, wouldn't the value of acceleration be entered as g or -g depending on which way up you happened to be working? The fact is that you have to be aware of the direction if you want the right answer. 9.81ms-1 is not enough.
    Perhaps I'm just saying that it could be confusing to treat g as positive, automatically and not be constantly aware of sign. But, if the 'm' in 'ms-1' is only distance and not displacement, by definition, then fair enough.
     
  21. Jul 11, 2012 #20

    Doc Al

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    Sure. The acceleration of an object in free fall would be, using your sign convention, -g.
    There's nothing in free fall in this problem. The only appearance of g in your problem is to give you the magnitude of the weight, mg.
    As long as you insist on using the equation N-W = ma you'd better learn what the symbols stand for. In that equation, W stands for the magnitude of the weight. The downward direction is already incorporated in the minus sign.

    Much better, in my opinion, is to learn the vector form of Newton's 2nd law as I describe in my first post. Then you'd really understand how things work.
     
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