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Newton's 3rd Law Consideration in Circular Motion

  1. Nov 29, 2011 #1
    1. The problem statement, all variables and given/known data

    The text-book question is, calculate the push of the road on the car with a mass of 900kg traveling with a speed of 18m s-1 over a hump-backed ridge.

    3. The attempt at a solution

    Push of the car on the road is equal to the push of the road to the car, hence:

    ƩF = Fc + W
    ƩF = (900x18/50)+(900x9.81) = 9153N

    I just want to confirm if my calculation and consideration is correct since I took it that the centripetal are also one direction with the weight of the car. In general I also want to confirm if we could just take Fc as ordinary force (push and pull) without any special care on them. It's because there's a similar question asking our effective weight considering the Earth centripetal force. I know intuitively that the centripetal force will cause our body to be actually accelerated away from the Earth hence reducing the gravitational attraction. Hence instead of adding the centripetal force to the weight force I subtract it.

    Effective Weight = mg - Fc?

    Could anyone clarify this both problems because they seemed to be contradictory?

    Thank You
     
  2. jcsd
  3. Nov 29, 2011 #2
    Let F = ma

    where F is the resultant (net) force and a is the centripetal acceleration.

    Now F is contributed by the normal reaction of bridge on car and by the pull of the Earth (weight) on the car.
     
  4. Nov 29, 2011 #3
    Is it possible then to write like what I've written?

    ƩF = Fc + W
     
  5. Nov 29, 2011 #4

    Doc Al

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    The way to solve for the normal force (N) between road and car without confusing yourself is to NOT think of Fc as a force on its own. The only forces acting on the car are the weight (which acts down) and the normal force of the road (which acts up).

    Now apply Newton's 2nd law, realizing that the acceleration is downward (towards the center).

    ƩF = ma

    choosing down to be positive:
    W - N = mv2/r

    Thus: N = W - mv2/r
     
  6. Nov 29, 2011 #5
    Ah so it is possible to say that the centripetal force is actually equal to whatever the resultant force of the body traveling in circular motion is? Whether if we replace the normal force with a tension of string for instance? Another way of saying is, the centripetal force is equal to the force that makes a body moves in circular motion?

    For example, is this a correct way to calculate the tension of a string at the top of a roller-coaster loop?

    W + T (since T has the same direction as W) = Fc
    T = Fc - W
     
  7. Nov 29, 2011 #6

    Doc Al

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    'Centripetal force' is just the name for the net radial force. It always equals mv2/r.

    Sure. I assume you mean the tension in a string whirling an object in a vertical circle at the top of the motion. (The top of a roller coaster loop would give a similar answer for the normal force.)

    What does this problem have to do with Newton's 3rd law?
     
  8. Nov 29, 2011 #7
    Ah ok, I'm just confused at first whether to take the action reaction pair of the centripetal force because I used to consider it as a real physical force similar to gravity or electrostatic force. Thanks for the clarification Doc Al.
     
  9. Dec 2, 2011 #8
    Quick question since I don't want to start a new topic.

    I've just found out that if we set the downward direction as negative then the resulting centripetal force will also be negative, is this desired? I mean, can the centripetal force be negative?
     
  10. Dec 2, 2011 #9

    Doc Al

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    Sure it can. Whether a force is negative or positive just depends on whatever arbitrary sign convention you have chosen. (The physics that matters is that centripetal acceleration is towards the center; whether you call that direction positive or negative is up to you. Just be consistent.)
     
  11. Dec 2, 2011 #10
    The simple way is to use F = ma in the direction of the centripetal force.

    Then the resultant (net) force F will be (W - N) where N will be a postive number since the direction of N is already taken care of by the minus sign.
     
  12. Dec 2, 2011 #11

    PeterO

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    Joining this discussion late:

    Particularly worried about your statement I have made red, above.

    Gravitational attraction force is determined by the mass of the earth, the Mass of the Car and the distance from the centre of mass of the car to the centre of mass of the Earth.

    Whether you are driving the car through a dip, over a hump-backed bridge or on flat ground, none of those quantities is appreciably changed, so the Gravitational Attraction is unchanged. We usually call that force the weight, and calculate its size using "mg". We also know the direction of that force [on the car] is towards the centre of the earth - often called "down".

    On flat ground, there is no vertical acceleration, so no net force. That means the Normal Reaction force is equal in size to the weight force, but is directed up. The weight force and the Normal reaction force cancel each other giving a net force of zero - and thus the net acceleration of zero. [remember this is only a vertical consideration].

    If the car is driven through part of a vertical circle [through a dip or over a humped backed bridge] there will be a resulting centripetal acceleration - achieved by there being a centripetal force. That force is directed toward the centre of the circle [of the motion].

    When driving through a dip, the centre of the circle is above the ground - often called up - so the net force is up.
    That is achieved by the Normal Reaction force becoming larger than mg.

    When driving over the humped back bridge, the centre of the circle is below the bridge - often called down - so the net force is down.
    This is achieved by the Normal Reaction force becoming smaller than mg.

    How does the Normal Reaction force manage to change size/strength? The key is in the word reaction. The surface reacts as required [or breaks trying] and supplies just as much force as is necessary.

    The real problem arises when so many people refer to the force as the Normal force - leaving out that key descriptor "reaction" which is the key to explanation.

    Examples:
    Place a 1kg book on a table - the table will push up with a force of 9.8 N , balancing the weight of the book - thus supporting the book.
    Place a 9 kg mass on top of the book: now the table must push up with a force of 98N to balance the weight of the couple.
    Add another 10kg - the table must push up with 196N
    Add another 10kg - the table must push up with 294N
    Keep doing this and we will reach a point when to support the book & masses, we are requiring the table to push up with a force larger than is possible. The table fails - though most people simply say the table breaks under the load.
     
  13. Dec 2, 2011 #12
    Alright, I've tried it myself and the answers are consistent. Thank You. :)

    Reading PeterO explanation, is it sensible to say that the only force our bodies can 'feel' is the normal reaction force then? I just realized the consequence of it now and why our perceived weight seems to be reduced when on a descending lift. It is very amazing what can one get from Newton's Laws.
     
  14. Dec 2, 2011 #13

    PeterO

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    Exactly - the ultimate reference being weightless when in free-fall, especially in an orbiting space capsule.
     
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