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Force of Gravitation and Centripetal Force

  1. Dec 14, 2016 #1
    1. The problem statement, all variables and given/known data
    (a) The Earth may be considered to be a uniform sphere of radius 6.37 × 103 km with its mass of 5.98 × 1024 kg concentrated at its centre. The Earth spins on its axis with a period of 24.0 hours. (i) A stone of mass 2.50 kg rests on the Earth’s surface at the Equator.

    1. Calculate, using Newton’s law of gravitation, the gravitational force on the stone.

    2. Determine the force required to maintain the stone in its circular path.

    (ii) The stone is now hung from a newton-meter.
    Use your answers in (i) to determine the reading on the meter. Give your answer to three significant figures.

    2. Relevant equations
    F= GMm/r^2 and F= mw^2r

    3. The attempt at a solution
    I got the solution for 1. 24.6 N and 2. 0.0842 N but I don't understand why is (ii) or the weight the difference between these two values? Why do we have to subtract Fg-Fc, aren't they supposed to be the same thing?
     
  2. jcsd
  3. Dec 14, 2016 #2

    CWatters

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    You have correctly calculated the gravitational force on the stone (24.6N) and the centripetal force required to maintain the stone in it's circular path (0.084N).

    As for your question...

    First thing to understand is that Fc is not a force that acts on the stone.

    The calculation Fc = mv2/r gives you the net force (eg the sum of all other forces) required in order for the stone to move in a circle of the required radius and velocity.

    There are two forces that act on the stone. One of the forces acting on the stone is Fg. What is the other force acting on the stone? Hint: Draw a free body diagram of the stone.
     
  4. Dec 14, 2016 #3
    The other force is the weight of the stone?
     
  5. Dec 14, 2016 #4

    cnh1995

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    Look up 'normal reaction'. It often shows up in the FBDs of this type of situations where you have one body placed on another body (or surface).
     
  6. Dec 14, 2016 #5
    Oh yes sorry! So how does that help us?
     
  7. Dec 14, 2016 #6

    cnh1995

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    It is one of the forces acting on the stone. Did you draw the FBD?
     
  8. Dec 14, 2016 #7
    Ok so the normal reaction is acting perpendicular to the surface of the earth where the stone has been placed and Fg is acting downwards and Fc (net force of the two) is inwards?
     
  9. Dec 14, 2016 #8

    cnh1995

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    Yes.
     
  10. Dec 14, 2016 #9
    But I still don't understand why weight = Fg - Fc?
     
  11. Dec 14, 2016 #10

    cnh1995

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    It's what is called the 'apparent weight' of the body. It is less than Fg (mg).
     
  12. Dec 14, 2016 #11
    I don't get it, why is it minus Fc? What did normal reaction have to do with this? How did you know for the last part of the question we had to calculate the apparent weight of the stone?
     
  13. Dec 14, 2016 #12

    cnh1995

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    The true weight of the stone, if the earth suddenly stopped rotating, would be simply Fg(=mg) and the normal reaction would be equal to the true weight. But on the rotating earth, in order to maintain the circular motion of the stone, the normal reaction is less than the true weight of the stone such that the difference between them, or the net force on the stone is the necessary centripetal force.
    I am not sure what the newton-meter should read. If the stone is suspended, it is no longer in contact with the surface, so no normal reaction. But there will be tension in the string to which the stone is tied , which will act same as the normal force. So the newton-meter should show the apparent weight.
     
  14. Dec 14, 2016 #13
    Ok that makes sense thanks.

    One last question, Fg and the normal reaction don't really act along the same line unless they're either at the top or the bottom of the circle. When they're on the sides, they make a 90 degrees angle with one another, then you can't do Fg-Fc can you? Because they're perpendicular to each other?
     
  15. Dec 14, 2016 #14

    CWatters

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    Ok step by step..

    The first thing to note is that the stone is accelerating. Therefore the net force acting on the stone is not zero.

    So we can draw our FBD..

    Stone FBD.jpg

    The two forces acting on the stone are Fg and the normal force FN so we can write that as...

    Fg + FN <> 0

    In order to move in a circle the net force must provide the centripetal force Fc (=mv2/r) so....

    Fg + FN = Fc

    Rearrange to give the normal force...

    FN = Fc - Fg

    The weight call it FW (as measured by the newton-meter) is the same magnitude as the Normal force but has the opposite sign so..

    FW = - (Fc - Fg) = Fg - Fc
     
  16. Dec 14, 2016 #15

    CWatters

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    I can imagine a situation where Fg and FN aren't parallel. For example a stone near a mountain would experience gravitational force that wasn't perfectly vertical. Not to scale!..

    Stone near mountain.jpg


    In that case you might need to think about other forces such as Ff the force due to friction.

    Your starting point would be as before, that the net force = Fc..

    FN + Fg + Ff = Fc

    (vector addition obviously)

    You then rearrange it in the same way as before to give FN and hence FW.

    FN = Fc - (Fg + Ff)

    FW = (Fg + Ff) - Fc

    Friction Ff can be calculated from the horizontal component of Fg
     
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