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Is support force a true force ? it seems incompatibe with F=ma

  1. Mar 22, 2013 #1
    Is "support force" a "true force"? it seems incompatibe with F=ma

    For example if a box is sitting on a table, not moving, it is said that the table exerts a support force on the box.

    But the box and the table are both inertial right? so if F = ma (force = mass x acceleration) and acceleration is defined as the rate of change of velocity, and velocity is defined as the rate of change of position of an object, but the table is stationary, then how could there be any velocity? and if there's no velocity than how is there any acceleration? and if there's no acceleration than how can there be any force? ( m * 0 = 0 ).

    Would it be proper to say that the table has no net force, if so could it still have a supportive force? is supportive force a real force in terms of F = ma or is it a not a "true force" sort of in the way centrifugal force is not a "true force"?
     
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  3. Mar 22, 2013 #2
    The box on the table feels a downward force (its weight) and an upwards force (the support from the table). The forces exactly cancel, so there in no net force and thus no net acceleration.

    The table has a downward force (its weight), another downward force (the weight of the box) and an upwards force (the support from the ground) which exactly cancels the two downward forces. There is no net force on the table, so there is no acceleration.

    The support force is there, it's very real, and it wants to accelerate the table. But it can't, since the weight of the table and box exactly cancel it.
     
  4. Mar 22, 2013 #3
    Centrifugal force IS a true force. Swing a weight around on a string - the force you feel is the centrifugal force, and it is the reaction to the centripetal force you apply to keep the weight rotating in a circle. Swing the weight fast enough and it will pull you over - this is clearly a true force.

    Put yourself in the place of the table - do you feel the weight of the box? Is this a real force?
     
  5. Mar 22, 2013 #4

    A.T.

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    "F" here refers to the net force : the vector sum of all forces acting on the object (box). The net force on the box is zero here:

    F_net = F_table_support + F_box_weight = 0
     
  6. Mar 22, 2013 #5

    A.T.

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    You are referring to the Reactive centrifugal force, which is indeed an interaction (real) force. The OP probably refers to the inertial (fictitious) Centrifugal force, in rotating reference_frames.

    There is some confusion of the two, and long discussion here on the forum. Here are the differences between the two:
    http://en.wikipedia.org/wiki/Reactive_centrifugal_force#Relation_to_inertial_centrifugal_force
     
  7. Mar 22, 2013 #6
  8. Mar 22, 2013 #7

    A.T.

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    There is no rotation at all in his example. He is not talking about some centrifugal force in his example, but just using "centrifugal force" as an example of "not a true force". So I assume he means the the "fictitious" centrifugal force, not the "real" reactive centrifugal force.
     
  9. Mar 22, 2013 #8
    Really? You think that while needing some reassurance in his (or her) understanding of reactive static forces he has sufficient grasp of vector mechanics to use components of its equations of motion as examples?

    Or has he just been confused by a combination of being told that the force you feel when on a roundabout is not centrifugal force but centripetal force (which is of course true, but is hardly ever accompanied by an explanation of things from the roundabout's point of view), and then by the fact that he has heard the statement "centrifugal force is not a real force" from those who are talking about something completely different?
     
  10. Mar 22, 2013 #9

    russ_watters

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    Short answer: f=ma is not the only cause/effect of force.
     
  11. Mar 22, 2013 #10

    stevendaryl

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    This disagreement happens all the time, and there seems to be no permanent resolution to it. But I have to disagree. When you are swinging a weight around on a string, the force you feel is not centrifugal force, it is tension in the string. You are pulling on the string, and the string is pulling back on you. Neither of those two forces is centrifugal force.

    At the other end of the string, the string is pulling on the weight, and the weight is pulling back on the string. Neither force is "centrifugal force".

    The only place where the concept of "centrifugal force" can possibly apply is right at the weight. The weight is being pulled inward by the tension force of the string, [itex]\vec{T}[/itex]. The centrifugal force is [itex]m \omega^2 \vec{r}[/itex] where [itex]\omega[/itex] is the angular velocity of the weight, and [itex]\vec{r}[/itex] is the vector from the center of rotation to the weight. These two quantities are equal and opposite:

    [itex]\vec{T} = - m \omega^2 \vec{r}[/itex]

    There are two different ways of interpreting the same equation:

    1. [itex]\vec{T}[/itex] is the force [itex]\vec{F}[/itex], and [itex]-\omega^2 \vec{r}[/itex] is the acceleration [itex]\vec{a}[/itex], and they are proportional by Newton's law: [itex]\vec{F} = m \vec{a}[/itex].
    2. [itex]\vec{T} + m \omega^2 \vec{r}[/itex] is the force [itex]\vec{F}[/itex], and [itex]\vec{a} = 0[/itex] is the acceleration, and they are proportional by Newton's law: [itex]\vec{F} = m \vec{a}[/itex].

    So you can consider the "centrifugal force" to be a force term and put it on the same side of the equation as the tension, or you can consider it to be an acceleration term, and put it on the other side of the equation.
     
  12. Mar 22, 2013 #11

    stevendaryl

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    I'm not exactly sure what your question is. The support force is certainly a real force. It doesn't produce any acceleration because there is a second force, gravity, that also applies to the box. Newton's law is [itex]\vec{F}_{net} = m \vec{a}[/itex], where [itex]\vec{F}_{net}[/itex] is the vector sum of all forces acting on the object. So not every force produces an acceleration, unless it is the only force acting on an object.
     
  13. Mar 22, 2013 #12
    So if you swing a solid rod around you feel centrifugal force but if you swing a weight/string system around feeling the same force it is something different? How can you distinguish between the two different forces with your eyes closed (don't try this at home...)?

    That is centripetal force.

    That is centrifugal force - or if you insist, reactive centrifugal force although as the concept of (reactive) centrifugal force was understood hundreds of years before vector mechanics was developed and some dimwit decided to call something else centrifugal force a distinguising prefix should be used when talking about THAT force if you don't want to go around confusing people who are not speaking in the context of vector mechanics.

    Here's a partial resolution that applies in this situation: if someone is describing a system using an inertial frame of reference it is reasonable to assume that when they talk about "centrifugal force" they don't mean something that only exists in a rotating frame of reference.

    Given the word "force" in "centrifugal force" and that there is no net acceleration arising in part from this force (it must be resisted by some equal and opposite force holding the centre of rotation in place otherwise there would be no rotation) I think it would be doubly peverse to consider it as an acceleration term. Do you consider centripetal force to be an acceleration term?
     
  14. Mar 22, 2013 #13

    Doc Al

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    Well, those "dimwits" have been running the show for many decades now, at least in physics. In modern physics usage, the term "centrifugal force" (without any modifier) means the inertial (non-interaction) "force" that one adds when viewing things from a rotating reference frame. Better get used to it!

    If you want to use the old-fashioned concept of centrifugal force, you'd better specify it as reactive centrifugal force. (But why even give it a special name?)
     
  15. Mar 22, 2013 #14
    To clarify, I was talking about inertial centrifugal force obviously.

    Back to the topic.....

    Assuming the supportive force is true (whether or not it is?) then what is the net force acting on the two items (the box and the table)?

    An above post says it is zero net force. That's what I assumed, but I was told elsewhere this is wrong so is there any basis for it being wrong?

    And in regards to it being a true force, what exactly is the conditions that qualify a force to be considered "true"? and which of these does it or doesn't it satisfy?
     
  16. Mar 22, 2013 #15
    If the net force is not zero, then something will be accelerating. Like you said, F=ma.

    It's a very easy experiment to perform. Look at the table....is anything accelerating? If not, then net force is zero :)
     
  17. Mar 22, 2013 #16

    Doc Al

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    Since they aren't accelerating, the net force on each must be zero.

    Who said it was wrong? (Were they nitpicking about the fact that since the earth is rotating, there is a centripetal acceleration? :rolleyes:)

    Well a "real" force can be traced to some basic physical interaction, such as an electromagnetic force or a gravitational force (in Newtonian physics). In other words, real forces have actors. The normal force of the table pushing up on the box is a real force; the table is the "actor" exerting the force. "Real" forces appear in every reference frame.

    Inertial forces have no "actor" since they are an artifact of viewing things from an accelerating reference frame.
     
  18. Mar 22, 2013 #17
    It seems to me that confusion is right at the beginning.
    In the equation F = ma, F is the RESULTANT (net) force. It is not any particular single force.
    You need to grasp this to have any chance of understanding what really is basic physics.
    Someone else has pointed out that F is the net force.
     
  19. Mar 22, 2013 #18
    Well centuries in fact! I totally accept what you say, but I still maintain that it is not helpful to introduce the concept of inertial centrifugal force before one has a firm grasp of Newtonian mechanics in general and rotational motion in particular. It certainly confused the hell out of me until I had the maths to cope with it, and it seems to be confusing another poster on this thread who seems to know the maths but has lost the concept of reactive centrifugal force altogether!

    Sorry for the hijack ihaveabutt.
     
    Last edited: Mar 22, 2013
  20. Mar 22, 2013 #19

    A.T.

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    The net force on the box is zero.
    The net force on the table is zero.
    The net force on the table+box is zero.

    See post #2
     
  21. Mar 22, 2013 #20

    stevendaryl

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    No, you never feel centrifugal force under any circumstances. It's not a real force. If you spin around a rod, then you feel tension forces in the rod.

    That name is a misnomer, and my opinion.

    The accounting of forces and accelerations for a weight spinning at the end of a string is:
    1. The weight is acted upon by an unbalanced force [itex]\vec{F} = \vec{T_1}[/itex] due to the string.
    2. The acceleration of the weight is radially inward: [itex]\vec{a} = - \omega^2 \vec{r}[/itex], where [itex]\vec{r}[/itex] is the vector from the center to the mass.
    3. By Newton's 2nd law, [itex]\vec{F} = m \vec{a}[/itex], so we have, for the weight: [itex]\vec{T_1} = - m \omega^2 \vec{r}[/itex]
    4. The string is pulled on by the weight, according to Newton's 3rd law. So the end of the string has an outward force [itex]\vec{F} = -\vec{T_1} = + m \omega^2 \vec{r}[/itex]
    5. The string also pulls on your hand with tension force [itex]\vec{T_2}[/itex]. If the mass of the string can be neglected, then this will be the same magnitude as [itex]\vec{T_1}[/itex], but in the opposite direction: [itex]\vec{T_2} = - \vec{T_1} = + m \omega^2 \vec{r}[/itex]
    6. By Newton's 3rd law, your hand exerts a force on the string that is equal and opposite to [itex]\vec{T_2}[/itex]. So it's a force [itex]-m \omega^2 \vec{r}[/itex]

    So you can account for all the accelerations and forces without mentioning centrifugal force, and in my opinion, it's confusing to even bring it up. The weight is accelerating, compared with the path of a "free" test particle, and that relative acceleration is the acceleration that is important for the application of Newton's laws.

    No. It's a force. In this case, it's the tension in the string.
     
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