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Newton's 3rd law and friction

  1. Apr 22, 2005 #1
    Hi everyone,

    I need some elightenment on one of the aspects of Newton's 3rd law. It's not a homework question but I'd like to gain some understanding of this concept.

    Let's say, a car accelerates from rest on a road. The tires pushes against the road (action force) and the road pushes back on the wheels (reaction force). I can't really seem to grasp this.

    Say the wheel exerts 5N of the road, while the road exerts 5N on the wheel. Wouldn't the forces cancel out? If the wheels produce a more force as the accelerator is being stepped on, wouldn't the reaction forces cancel out as well resulting in 0 acceleration?

    My thoughts are that when the wheel exerts force on the ground, the earth exerts force on the wheel. And the earth accelerates as well, but it's not noticeable because of the huge mass of the earth but the car has a much lower mass resulting in a noticeable acceleration. Is this right? Haven't been able to convince myself on this as I'm still a blurred by the cancelling out of the forces.


    How do you determine what is the maximum friction a surface could produce?

    Sorry for the long post, thanks a lot :)
     
  2. jcsd
  3. Apr 22, 2005 #2

    arildno

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    You have some basically correct ideas here, but you have them mixed up with another issue; which is why you are confused on this.

    I'll see if I can clear the matter up for you:
    Suppose you've got two objects, 1 and 2, and 1 exerts a force called [tex]\vec{F}_{12}[/tex] on object 2, whereas object 2 exerts a force on object 1, called [tex]\vec{F}_{21}[/tex]
    Their masses and accelerations are called [tex]m_{1},\vec{a}_{1},m_{2},\vec{a}_{2}[/tex] respectively.

    Thus, this being the only forces present, Newton's 2. law for the objects is separately:
    [tex]\vec{F}_{21}=m_{1}\vec{a}_{1}[/tex]
    [tex]\vec{F}_{12}=m_{2}\vec{a}_{2}[/tex]
    Are we in agreement thus far?

    Now, Newton's 3.law simply states: [tex]\vec{F}_{21}=-\vec{F}_{12}[/tex]
    That is action and reaction are equally big, but oppositely directed and, most importantly they act upon DIFFERENT objects.
    Now, if we drop the subscripts from the force, and also the vector arrow, we find for accelerations:
    [tex]a_{1}=\frac{F}{m_{1}}, a_{2}=-\frac{F}{m_{2}}[/tex]
    Agreed?

    But let us now consider the two following issues (and it is here your confusion lies, I think)
    1) Does the distance between the objects change?
    2) Does the CENTER OF MASS of the two objects experience any acceleration as a result of the force couple?

    The answer to 1) is yes:
    If l(t) is the distance between the two objects, then [tex]a(t)=a_{1}-a_{2}[/tex] measures how fast object 1 accelerates away from object 2.
    From the above, we have:
    [tex]a(t)=F\frac{m_{1}+m_{2}}{m_{1}m_{2}}\neq0[/tex]

    Thus, the distance between two objects influencing each other through a force couple will then, in general, change.

    But, the answer to 2) is no.
    The acceleration of the center of mass is given, by definition through the equation:
    [tex]a_{C.M}=\frac{1}{m_{1}+m_{2}}(m_{1}a_{1}+m_{2}a_{2})=\frac{1}{m_{1}+m_{2}}(F+(-F))=0[/tex]
    where I used the info from Newton's 2. and 3.laws to derive the result.

    Thus, what a action/reaction couple "cancels out" is the acceleration of the two objects COMMON CENTER OF MASS (it doesn't provide any to begin with).

    I hope this helped..
     
    Last edited: Apr 22, 2005
  4. Apr 22, 2005 #3
    Oh I just saw arildno posted your answer. Cheers!
     
  5. Apr 22, 2005 #4

    OlderDan

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    How about I start you on the first one, and leave the rest to somebody else or a later time.

    The answer to your question is in your own words.

    "Say the wheel exerts 5N ON THE ROAD, while the road exerts 5N ON THE WHEEL."

    These two forces act on different objects. Forces only cancel when they are acting on the same object, (or a group of objects considered to be one object subject to forces external to that group). The 5N force acting on the road can only be canceled by some other force acting on the road, not by a force the road exerts on the wheel. The 5N force acting on the wheel can only be canceled by some other force acting on the wheel, not by a force the wheel exerts on the road.

    If you can grasp the idea that action-reaction always act on different objects, and that such forces cannot cancel one another, you will be well on your way to clearing up other difficulties.

    Edit- Late to the party again I see :rolleyes: I'll leave it hoping it is of some value
     
    Last edited: Apr 22, 2005
  6. Apr 23, 2005 #5
    Hello,

    Thanks very much for the explanation. Indeed great help, thanks again :). Do you mind elaborating a bit more on how you used Newton's 2nd and 3rd law to derive that equation for question 2? Thanks! :)
     
  7. Apr 25, 2005 #6

    OlderDan

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    Consider two objects exerting a force on one another and under the infulence of external forces. Let's assume one dimensional motion, but all this can be extended to three dimensions. Newtons second law tells us that

    [tex]F_{e1} + F_{i1} = m_1a_1[/tex]

    [tex]F_{e2} + F_{i2} = m_2a_2[/tex]

    Newton's third law tells us that the two "internal" forces are equal and opposite. If we add the two equations, those forces add to zero.

    [tex]F_{e1} + F_{e2} = m_1a_1 + m_2a_2[/tex]

    If an object whose mass is the sum of the two masses were acted upon by the total force external to the two objects it would experience an acceleration. We call that the acceleration of the center of mass of the two objects.

    [tex]F_{e1} + F_{e2} = (m_1 + m_2)a_{CM}[/tex]

    so

    [tex]m_1a_1 + m_2a_2= (m_1 + m_2)a_{CM}[/tex]

    The CM designation comes from the definition of center of mass, which is

    [tex]m_1x_1 + m_2x_2 = (m_1 + m_2)x_{CM}[/tex]

    Taking one derivative with respect to time of this equation
    gives

    [tex]m_1v_1 + m_2v_2 = (m_1 + m_2)v_{CM}[/tex]

    and taking another derivative gives the equality involving accelerations above. Solving for the acceleration of the center of mass we get

    [tex]a_{CM} = \frac{m_1a_1 + m_2a_2}{m_1 + m_2}[/tex]

    In the absence of any external forces the two objects can accelerate, but by Newton's third law the first two equations lead to

    [tex]m_1a_1 = - m_2a_2[/tex]

    This means that the acceleration of the center of mass is zero. It also implies that

    [tex]m_1v_1 + m_2v_2 = constant[/tex]

    In other words the total momentum of the two particles cannot change in the absence of external forces. Momentum must be conserved unless external forces are acting. It is often convenient to look at the motion of two objects from a center of mass frame of reference. From that perspective

    [tex]m_1x_1 + m_2x_2 = 0[/tex]

    [tex]m_1v_1 + m_2v_2 = 0[/tex]

    The total momentum in the center of mass frame of reference is zero.
     
  8. Apr 25, 2005 #7
    Thanks Olderdan really cleared things up for me
     
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