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Homework Help: Clarification of the normal force

  1. Jul 31, 2010 #1
    Hi, I'm a new member here! I don't like introductory by questions, but I think this is the best place to do it. My questions concern the normal force, Something that seems intuitive at first, but the details a little unclear.

    1. The problem statement, all variables and given/known data
    I'll break down my questions into parts:

    1) I'm taught in my high school syllabus that a weighing scale (weight) measures the normal force on an object. My first question concerns on how exactly is it really measured?

    Of course, the normal force is essentially... a force. Therefore, I infer that the scale utilizes some "springy" mechanism, and some kind of system measuring how much it is "resisting" the force on it. Is this so? If it is, then on to my second question..

    2) I wanna illustrate this with a classic textbook example: measuring the "apparent weight" of an object A on an accelerating elevator using a weighing scale.Now let the elevator be accelerating upwards. I am taught that the Normal force (read from scale) = mg + ma (I think this formula accounts only for the magnitude [or converts the mg downwards to it's third law pair), thus no need for negative values).

    So my 2nd question is how exactly does the scale measure this? I can't see a way how my hypothesis for the 1st question explains this: I can't see how the spring "resists" the force upwards while adding up to the normal force that way. The whole scale accelerates upwards, and I don't see some parts of the spring accelerates more than the other parts of the spring. I'm probably mistaken something here.

    3)Now, take the normal force, mg + ma. Newton's third law of motion asserts that there is an opposing force exerted by the object on the scale in the direction of gravity. My question for this part is this: How is this possible when the object is accelerating upwards? Where did the additional force ma come from? I pose a similar question when the elevator is accelerating downwards. The scale reads mg - ma. But if the body (and everything) is accelerating downwards, why is the force on the scale reduced by ma? Clearly, I am misunderstanding something. What is wrong with my argument?

    I think I get it if I see it in a free body diagram of the situation, when the vectors superposes and adds up, but I'm missing something in the details in how the "flow" of the force works. A "chronology" of the happenings will help here, I think, like the accelerating elevator will cause A to happen and then A will cause B to happen. I think essentially, my problem lies on how and why the vectors (counter gravitational pull force + the force accelerating the body) adds up. Or is it just a "detached happening" just on the scale?

    4) Suppose an object A is at rest on a flat weighing scale on earth. Newton's third law of motion asserts that there is an opposing force of gravity which is equal in magnitude but opposite in direction. So Normal force = mg.

    Now suppose the object A is free-falling and strikes the same flat weighing scale. According to Prof. Julius Miller in his program, the scale reads twice the weight of the object during contact. In this situation, it seems that the Normal = mg (gravitational pull) + ma (the push on the scale? the net force? Or is it contact force? Not so sure here too.). How exactly does the vector add up? Should the vectors of different "type" of force superpose like that or am I misunderstanding what Prof. Miller is trying to explain?

    I am aware that the gravitational pull and the normal force are not third law pairs, but I think the problem I'm having is with the understanding of the normal force's counter pair (contact force?). I hope I stated my question clearly.

    Thanks in advance!
  2. jcsd
  3. Jul 31, 2010 #2

    Doc Al

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    The scale just reads the force acting on it. (It actually reads the displacement of some spring--depending upon the details of the mechanism--which is then displayed in terms of force.)

    Since the object is sitting on the scale, do you not agree that the scale must be pushing up on the object? And since everything is accelerating upwards, you can deduce that the normal force exerted by the scale must exceed the weight (mg) by an amount equal to ma.

    I'm not sure how this question differs from the previous one.

    Careful! The normal force is not the 'third law pair' to gravity. Gravity is an attraction of the earth on the object; in turn, the object exerts a gravitational force on the earth.

    The normal force is a contact force that the object and scale exert on each other.

    The normal force is a contact force.

    See my comments above.

    I'm not sure if my answers will provide any help, but please keep asking.
  4. Jul 31, 2010 #3


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    Start out from Newton's second law F=ma and read it in the opposite way: If a body of mass m accelerates with "a" than it experiences the resultant force F=ma.

    You stand on a weighting scale in rest. a=0, the resultant force is F=0. But you know that the Earth pulls you downward with the force G=-mg.
    The scale exerts the upward normal force, N. F=N-mg=0, so the scale must act on you with a force equal to your weight: N=mg.
    This force is supplied by a spring inside the scale which is pressed by you and shortened relatively to its relaxed length. You know Hook's law: The spring force = -kΔL. The readout of the scale is proportional to ΔL.
    You press the spring, the spring presses your feet. (Yes, you do not stand on the end of the spring but on a flat plate attached to it, so you press this solid plate, it presses the spring, and it goes on also on the opposite way.)
    If you are in a lift which accelerates upward, you accelerate together with it (after a very short time). If the acceleration is a, the resultant force is F=ma. Again, this resultant force comes from gravity G=-mg and the upward force of the scale N: ma=N-mg. So the force the scale exerts on you is N=ma+mg=m(a+g) The spring force is bigger, than it was in rest, the spring has to be shortened more, you read a bigger weight on the scale.

    In the unfortunate case when the suspension is broken and the lift is in free fall, you fall with it with acceleration a=-g. F= -mg. F=N-mg = -mg, so N=0. The scale do not push you upward, you do not push the spring, it has its equilibrium length so the scale reads zero weight.

  5. Jul 31, 2010 #4


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    You made me remember of a high school exersice where it was asked to prove that for a body at rest on a table , the gravitational force (body's weight) is equal to the force that the body exerts on the table.

    The normal force is the force that the table exerts on the body and due to newton's 3rd law the body exerts an equal but opposite force on the table. Since body is at rest weight+normal force=0 hence normal force=-weight hence (force from body to table)=-normal force=weight.
  6. Jul 31, 2010 #5
    The mechanism of a spring weighing scale is pretty simple. When you put an object on a pan, (at the very beginning) the only force acting on it is gravitational. Then, as the body slowly goes down, the pan is acting on the body with a normal force in the opposite direction. At a certain moment, the whole thing stops moving, and the equilibrium is established. Now, it is important which forces act on which object here.

    Object of measurement: (1) gravitational force (Earth) [tex]\downarrow[/tex], (2) reaction (normal) force (pan) [tex]\uparrow[/tex]
    Pan: (3) reaction force (object) [tex]\downarrow[/tex], (4) elastic force (spring) [tex]\uparrow[/tex]
    Spring: (5) reaction force (pan) [tex]\downarrow[/tex], (6) reaction force (force on the other side of the spring) [tex]\uparrow[/tex]

    Each force is numbered, and the origin of each force is written in parentheses. Directions are also indicated. Forces (2) and (3) are the pair of forces from the 3. Newton's law and so are (4) and (5). All forces are equal by magnitude (because nothing moves).

    I hope you understand the mechanics of this type of scale now and why we say that scale measures the normal force.

    Let say that we have a spring attached to the ceiling of the elevator. That will be our scale. If there is a force pulling the elevator upwards, then the same force acts on the scale (spring) too. But, there is an object attached to it on the other side pulling the spring to the opposite direction. Now, let's see the forces.

    Body attached to spring: (1) gravitational force (Earth) [tex]\downarrow[/tex], (2) elastic force (spring) [tex]\uparrow[/tex]
    Spring: (3) reaction force (body) [tex]\downarrow[/tex], (4) "upwards" force (origin doesn't really matter) [tex]\uparrow[/tex]

    If you look from outside the elevator you see the body accelerating, so you have equation (2. Newton's law)

    [tex]m\vec{a}=m\vec{g}+\vec{F} \Rightarrow F=m(a+g)[/tex],

    where [tex]\vec{F}[/tex] is elastic force (shown on the scale). Just try to draw the force diagram for the body attached to scale and everything will be clear.

    If an object falls on the scale, it will still show the same weight after some time (when the spring stops oscillating). During the oscillation, the scale shows different values, depending of the height from which body falls. I don't see why would measured weight be twice the weight of the object, although it could happen only at the certain moment(s).
    Last edited: Jul 31, 2010
  7. Jul 31, 2010 #6

    D H

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    The acceleration when you are standing at rest is zero -- but only in a rotating frame of reference. From the perspective of an Earth-centered non-rotating frame of reference, i.e., an Earth-centered inertial frame, a person standing on the surface of the Earth (poles excluded) is accelerating. A scale with 0.1 lb accuracy is more than accurate enough to be sensitive enough to this effect.

    That is a bit nit-picky, but then again the original poster has been correctly taught that a spring scale measures the normal force on an object. To OneWD: Suppose you climb a step ladder with scale in hand. Put the scale against the ceiling and push up on it. The scale will register a non-zero reading. That doesn't mean that gravity has somehow reversed direction. The scale isn't measuring gravity. It is measuring the normal force on the face plate of the scale. (Even more precisely, it is measuring the compression of a spring.)
  8. Aug 1, 2010 #7
    Thanks for the analysis! Clears up much of my confusion. I have a few questions though.

    1) Are forces (1) and (6) third law pairs here? Can I approximate it as the center of gravity of earth (earth is pulled by the object but because of the excessive mass does not accelerate) or does it go by the same reasoning you apply on the pan, where the other part of the spring is "resisting" the force of gravity? I'm not comfortable with my approximation though. It's probably wrong.

    2) This is just a thought that seemed to me wrong, but I have no idea how to rebut it. It's probably a lack of understanding of vectors. The problem goes like this: In a simple case where an object A rest on a flat surface on earth, the force of Gravity is pulling the object A towards the surface, force = mg. The surface prevents the object falling in by exerting a force of -mg on the object. By Newton's third law of motion, we deduce that there is a reaction force mg acting downwards by the object onto the surface. But it seems now that the surface now "feels" a force of 2mg! Why can't I add the vectors this way?

    Thanks for all the replies on this! I've noticed the clumsiness in my reasoning for question number 2 and 3. But the essential question of it still remains which I didn't think I stated directly. I understand that the normal force must equal to the m(g + a), but according to Newton's third law, the body will equally exert the same force downwards, i.e -m(g + a). But then, wouldn't the scale be in equilibrium? But actually it's accelerating! What is flawed in my argument?

    Thanks for that bit of information! I think that piece is vital, too bad my syllabus did not cover it.

    Okay, I think I'm getting it. I'll write down what I think I've learned here:

    1) The normal force acts before the contact force. (In this particular scenario, when the only force acting on the body is the force of gravity. I think there are other cases when the contact force comes first i.e. a collision with a moving object.)

    2) Free body diagrams only works when referring to the exact force(s) acting on the object that is being observed. (This seems obvious, but from what I've learned from my text {where it only supply vector arrows and not much on what the actual vectors mean, i.e. the mix up I have with gravity and the contact force} it is good reinforcement.

    And one that I need some confirmation on:
    3) There is no need for an object to be in motion (or acceleration) to exert a force on an object. A force causes acceleration, but not necessarily the other way. [Collision is an exception?]

    ... And a question still remaining:
    About the free-falling object striking a scale: At that instant, is the normal force equal to -mg + the impact force or does the impact force takes into account the -mg?

    Last edited: Aug 1, 2010
  9. Aug 1, 2010 #8
    No, forces (1) and (6) are not third law pair. A pair of forces exists between two different bodies acting on each other. So, as the Earth pulls the object "down", the object also pulls the Earth "up" with the same force - that's the force creating a pair with (1). Of course, the Earth also accelerates, but with a very small acceleration (because of the mass much greater than the body's). Maybe you didn't notice, but that's why I specified the "origin" of each force. It makes easier to identify third law pairs.

    The underlined sentence is completely correct! And that's the only force acting onto surface (from the body). The surface feels just the mg force, not 2mg. You must be careful when adding vectors - you can only add vectors of forces acting on the same body!

    Don't forget that m(g+a) is not the only force acting on the scale. As the scale is accelerating, it can't be in equilibrium. You tried to add forces acting on two different bodies, but that doesn't really make sense!
  10. Aug 1, 2010 #9

    D H

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    Last edited by a moderator: May 4, 2017
  11. Aug 1, 2010 #10
    I think the source of confusion is the action-at-a-distance "type" of force gravity is that is making me think that way. If the force of gravity only acts between earth and the object studied, I guess the only way we know that there exist a normal force which is equal in magnitude is because the net force on the object is 0. Is it true?

    What exactly is the force then? the scale "feels" a normal force of mg+ma on it, so the "upward force" must equal to at least (M+m)(g+a)? Sorry, I'm a little confused.

    EDIT: I didn't notice D H's latest post after I publish this. I'll go check out the links. Thanks D H!

    EDIT 2: I've made a mistake. Following my train of thought before, it shouldn't be m(g+2a), but (m+M)(g+a) where m= mass of object, M=mass of spring.
    Last edited: Aug 1, 2010
  12. Aug 1, 2010 #11
    Yes, that's the way we can clearly see that gravity is not the only force acting on a body lying on the surface (the body is at rest, so the net force must be 0). The real nature of normal force is electromagnetic (interaction between atoms, molecules etc.).

    Our point of interest is the spring, i.e. the forces acting on it. We have force m(g+a) acting downwards and another force F acting upwards. Since the spring accelerates (if you look from outside the elevator), F is greater than m(g+a). The F force depends on the mass of the spring.


    where M is the mass of the spring.

    But, the force we actually measure is the elastic force of the spring. It's the force that opposes stretching. If we want to stretch the spring, we need two forces in different directions. Here we have F and m(g+a). So, the spring actually "opposes" just the m(g+a) force, since F is greater and counterparts m(g+a).
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