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Can force be a scalar?

  1. Apr 7, 2009 #1
    If force is always parallel to displacement and displacement is always in a straight line and doesn't change direction. Then the displacement is equal to distance right?

    Distance is a scalar quantity. Does this mean that force parallel to displacement or distance as it would be is a scalar quantity?
     
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  3. Apr 7, 2009 #2

    Doc Al

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    No, but the magnitude of the displacement will equal the distance (in that special case). And it has nothing to do with the force.

    No. Force parallel to displacement is still a force, which is a vector.
     
  4. Apr 7, 2009 #3

    HallsofIvy

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    IF you have a "one dimensional" situation where you KNOW that everything, force, acceleration, velocity, position, lie in a straight line, then you can ignore the (fixed) direction and treat force, acceleration, velocity, postion, etc. as if they were scalars, using only the magnitude. It would be wrong to say they are scalars because you are always keeping in mind the fixed direction.
     
  5. Apr 8, 2009 #4
    Thank you for your help. I am also trying to connect Kinetic energy and Force times displacement. The problem I'm having is that I can only find

    change in work done = Force times change in displacement

    and

    Total work done = change in Kinetic energy

    by work-energy theorem

    What is need is

    change in work done = change in Kinetic energy

    is this still true

    I can therefore say

    change in Kinetic energy = Force times change in displacement

    dT = Fdr

    Is this true and if so is it also true for a variable force?
     
  6. Apr 8, 2009 #5

    Hootenanny

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    I'm guessing that this is in reference to your other thread where the force is a function of velocity.

    Assuming that we are in rectilinear motion (i.e. the force has one component parallel to the displacement and the displacement in a straight line) and that F=F(v),

    [tex]\begin{aligned}
    W & = F\cdot r \\
    \frac{dW}{dr} & = \frac{dF}{dr} + F \\
    & = \frac{dF}{dv}\frac{dv}{dr} + F \\
    & = \frac{dF}{dv} \frac{dv}{dt}\frac{dt}{dr} + F \\
    & = \frac{dF}{dv}\cdot a \cdot \frac{1}{v} + F
    \end{aligned}[/tex]

    So, the rate of work being done with respect to displacement is governed by the differential equation

    [tex]\frac{dW}{dr} = \frac{a}{v}\frac{dF}{dv} + F[/tex]

    Do you follow?
     
  7. Apr 8, 2009 #6
    Yeah I follow but it's not really on the same topic sorry. I have the relation between Work and force times displacement.
    It's simply

    delta W = Force (s) times delta s

    I need to know if

    delta W = delta K.E.

    I can therefore say

    delta K.E. = Force (s) times delta s

    but so far all I can find is

    Total W = delta K.E.

    Thanks again for you time

    sorry where delta is change in (difference)
     
  8. Apr 8, 2009 #7

    Hootenanny

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    Yes, in the absence of a potential, the work done on an body is equal to the change in kinetic energy of that body (ignoring the transformation into heat etc.).
     
  9. Apr 8, 2009 #8
    apologies for wasting your time I have jsut looked throught the notes I have and the reason I am confused is they say two different things. One book says

    delta W = force times delta r

    and the other says

    W = force times delta r

    do you know which is correct?
     
  10. Apr 8, 2009 #9

    Hootenanny

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    The former is correct. In words, "the infinitesimal work done by a force is given by the product of the force and the infinitesimal displacement in the direction of the force".

    No need to apologise!
     
  11. Apr 8, 2009 #10
    This brings me back to my original question

    can I say

    in the absence of a potential, the CHANGE IN work done on an body is equal to the change in kinetic energy of that body (ignoring the transformation into heat etc.).

    p.s what is a PF mentor?
     
  12. Apr 8, 2009 #11
    Whenever you speak something about direction, its a vector...And the word parallel denotes nothing but direction. So, still force is a vector quantity..:tongue:
     
  13. Apr 8, 2009 #12

    Hootenanny

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    The "change in work done" isn't entirely correct. Roughly speaking, the work done is the energy transferred to a body by a force acting through a distance. So your statement above would effectively read "the change in the energy transferred to a body...", which isn't correct.

    Effectively, you want to know the energy transferred to the object through the application of a force - this is precisely the definition of work as I stated above. Therefore, one could say "the work done on a body is equal to the change in kinetic energy ..."

    Does that make sense?
    Take a look https://www.physicsforums.com/library.php?do=view_item&itemid=88".
     
    Last edited by a moderator: Apr 24, 2017
  14. Apr 8, 2009 #13

    Doc Al

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    This is always true, if understood to refer to infinitesimals (as Hootenanny stated). Better to write it as:

    [tex]dW = \vec{F}\cdot d\vec{r}[/tex]

    In certain special circumstances, such as you describe in your first post ("If force is always parallel to displacement and displacement is always in a straight line and doesn't change direction."), then it is also true that W = FΔr. (Note that this is just the integral of the first formula as applied to this special case.)
     
  15. Apr 8, 2009 #14
    brilliant you guys are the masters! That was what I was looking for thank.
    If you have time could you take look a section of my report deriving drag and tell me what you think?
     

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