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Apparent weight

  1. May 25, 2014 #1
    In my physics book the equation for apparent weight is given as

    FN = mg + ma

    where FN is the normal force, m is the mass of the object, g is the gravitational acceleration of the object (= 9.8 m/s2) and a is the acceleration of the system.

    For example the system could be someone standing on a scale inside an elevator. Let's for simplicity's sake say this person weighs 100 kg. Let's also for simplicity's sake only consider the cases where the elevator goes from rest and accelerates either upwards or downwards, and not the cases when it is already moving and slows down. When the elevator is standing still then, the acceleration a of the system is 0, so the normal force will be FN = 100*9,8 = 980 N. The normal force is the apparent weight that is shown on the scale and when the elevator is standing still it's equal to the true weight.

    Now if the apparant weight is, let's say a) 700 N and b) 1100 N, how large is the acceleration a of the system? According to the equation above it will be

    a = FN/m - g

    so we will get

    a) a = 700/100 - 9.8 = -2.8 m/s2
    b) a = 1100/100 - 9.8 = 1.2 m/s2

    Now what puzzles me here is the sign of the acceleration. We know that the apparent weight is larger when the elevator goes from rest and accelerates upwards, and less when it goes from rest and accelerates downwards. Hence, the negative sign in a) implicates an elevator going from rest accelerating downwards.

    But if a = -2.8 m/s2 implicates something accelerating from rest DOWNWARDS, then why shouldn't the formula have g = -9.8m/s2, when g being gravitation obviously also means acceleration from rest downwards. It doesn't, in the equation we used a positive value for g! If we had used g = -9.8m/s2, both case a) and b) would have given the same sign which is incorrect.

    Where is the error? Why doesn't g and a, when both are going in the same direction (from rest to acceleration downwards) have the same sign?
     
    Last edited: May 25, 2014
  2. jcsd
  3. May 25, 2014 #2
    actually the formula seems a bit of no use to me.I always sketch the free body diagram of lift in my brain and then doing some pseudo force work the question is solved ALWAYS!
    in the first part for sure lift accelerate downwards. the guys who done with the formula considered "a" (the base of formula) in upward direction. therefore the acceleration is negative but negative with respect to their chosen direction of acceleration.
    work on your imagination and consider this as lift please see the FBD




    l---------------l
    l---------------l a[itex]\uparrow[/itex]
    l---[itex]\downarrow mg[/itex]-------l
     
  4. May 25, 2014 #3
    Don't you mean the other way around? If we draw a free body-diagram and have FN upwards and mg and ma both pointing downwards then we get exactly the equation from my book. If we had instead drawn ma upwards, the answers to a) and b) would have reversed signs.

    But assuming ma upwards also doesn't make sense to me. I mean, why should we assume mg and ma as pointing in DIFFERENT directions in order to get the same sign for acceleration in the same direction?

    I mean the way I think of it, logically it should be: same direction <=> same sign. So why isn't it like that here?
     
  5. May 25, 2014 #4

    Doc Al

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    g is just a constant; it's always positive.
     
  6. May 25, 2014 #5

    Doc Al

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    The speed of the elevator is irrelevant. All that matters it its acceleration. If it accelerating upwards, then the elevator floor is pressing harder into your feet to accelerate you will the elevator--thus the increase in your apparent weight. If it accelerates downward, then it presses less and your apparent weight is less.
     
  7. May 25, 2014 #6
    actually it all depends on sign convention
     
  8. May 25, 2014 #7

    Doc Al

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    Actually, it doesn't.
     
  9. May 25, 2014 #8
  10. May 25, 2014 #9
    I understand all that. The only thing I don't understand is the sign (+/-) of the acceleration. We have a free body diagram where we have assumed both mg and mg to be pointing downwards. Still, accelerating in the same direction they have opposite signs - even though they point in the same direction. Why? Same direction should be same sign!
     
  11. May 25, 2014 #10

    Doc Al

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    Most standard textbooks use the convention that g is just a constant equal to 9.8 m/s^2.

    Thus, if you choose a sign convention such that up is positive the acceleration of a falling object is: a = -g.
     
  12. May 25, 2014 #11

    Doc Al

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    Can you rephrase your question?

    On a free body diagram, the weight always points downward. So if up is positive, the weight is -mg.

    The acceleration could be positive or negative, depending on the situation.
     
  13. May 25, 2014 #12
    Ok, so we have a free body-diagram with FN pointing upwards and both mg and ma pointing downwards. So ma and mg are assumed to point in the same direction.

    Let's make it as simple as possible and ONLY consider the following scenario: The elevator is at rest, and then starts accelerating DOWNWARDS.

    Now, according to the equation, the normal force will be SMALLER during acceleration downwards than what it is when the elevator is at rest. So we can calculate the acceleration a by rearranging the equation:

    a = FN/m - g

    Now, when the normal force is smaller than before, the value of a will be NEGATIVE. SO we can conclude that a negative value of a corrsponds to acceleration DOWNWARDS.

    Now the question: g is ALSO acceleration DOWNWARDS. But g is positive, not negative like a. But if a and g are pointing in the same direction, WHY DO THEY HAVE OPPOSITE SIGNS?
     
  14. May 25, 2014 #13
    the a i made is acceleration of lift. so it's pseudo force acts downwards and equation comes out same. also 'a' is negative ,and the positive direction of 'a' is upwards so 'a' overall in example one NET ACTS DOWNWARDS. don't match it with sign convention because my explanation itself has an convention that positive direction of a is upwards.
    the pseudo force of ma acts downward as you want

    if one really want a formula he need to decide direction of acceleration, the makers did the same by choosing their direction as upwards. think over it
    if you feel me to be wrong then you must consider reverse situation i.e. acceleration downwards and design a formula yourself and you see value of 'a' positive in example 1
    new formula

    F=mg-ma
     
    Last edited: May 25, 2014
  15. May 25, 2014 #14

    Doc Al

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    On a free body diagram, only the forces FN and mg appear. mg always points down.

    OK, that means the acceleration is downward.

    OK.

    Again, g just stands for the magnitude of the acceleration due to gravity.

    Let's apply Newton's 2nd law: ∑F = ma

    First find the net force. The only forces acting are gravity (down, thus negative) and the normal force (up, thus positive).

    So: ∑F = -mg + Fn

    Set that equal to ma:
    -mg + Fn = ma

    So, when the normal force is less than mg, the acceleration is negative. Works out just fine.
     
  16. May 25, 2014 #15

    Doc Al

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    It's not necessary to introduce the concept of pseudo forces here. That's an advanced topic that is usually not covered at this level.
     
  17. May 25, 2014 #16
    But if we switch direction of ma like you are suggesting then ma and mg are pointing in different directions on the free body diagram. If they are pointing in different directions they should have opposite signs. I still don't understand why the directions of ma and mg doesn't give the same sign for the same direction?

    Why doesnt ma appear in the free body diagram?
     
  18. May 25, 2014 #17

    Doc Al

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    Only forces appear in a free body diagram. "ma" is not a force.
     
  19. May 25, 2014 #18
    What is it then?

    And how can the equation FN = mg + ma be made from the free body diagram if ma doesn't appear on it?
     
  20. May 25, 2014 #19

    ZapperZ

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    There is a severe misunderstanding of "F=ma" here.

    On the left hand side are all the forces acting on the object. The "resultant" goes on the right hand side of the equation, i.e. it resulted in an acceleration of the object, the dynamical result, the motion, etc. If the object doesn't move, this is where you equate all the forces on the left hand side to zero, such as the case when you deal with statics.

    Zz.
     
  21. May 25, 2014 #20
    Actually I disagree, I think if pseudo forces would explain this better, I would welcome this explanation.

    I think this pseudo force made it more understandable. What we have here is a non-inertial refrence frame which is fixed to the accelerating elevator system (including the scale but not including the person standing on the scale). So that's why ma is not considered a force acting on the scale, because it's the reference frame itself that is accelerating.

    So if we have this accelerating reference frame, when the elevator accelerates downward, it could be seen as that the gravitational force actually becoming less (in relation to the reference frame). And the FN is only the reaction of the gravitational force, so it also becomes less.

    Then for some reason, the fact that ma is the same as reference frame, makes it's signs be inverted. I still don't understand why this happens though. If someone could explain this particular detail, that would be great.
     
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