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Rate of acceleration

  1. Nov 16, 2007 #1

    IMK

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    Hello,

    Sorry to bother you all with this simply question but I could not think of a better place to ask.

    If a body/particle is travelling along an incline of 45 degrees from the horizontal at a constant speed, it is as I understand it accelerating away from the centre of the earth.
    So how do I calculate the rate of acceleration please?
    Many thanks IMK
     
  2. jcsd
  3. Nov 16, 2007 #2

    Doc Al

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    Staff: Mentor

    If the particle is maintaining a constant direction and speed, why would you think it's accelerating?
     
  4. Nov 16, 2007 #3

    berkeman

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    Staff: Mentor

    ..... and if it is going "down" the incline, why would you think it is moving "away" from the center of the Earth?
     
  5. Nov 17, 2007 #4

    IMK

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    Hello Doc Al, berkeman and many thanks for your reply

    Now as I understand it, and I am asking if I am correct?


    If I am travelling vertically away from Earth and a constant speed, my sense of gravity would be 1g plus the ng effect of the speed I am travelling at. Likewise if I was travelling towards Earth vertically at constant speed, my sense of gravity would be 1g less the ng effect of the speed I am travelling at. (This assume I am only a short distance from Earth and hence the affect of gravity is constant).

    So what I am asking is: I f I am travelling at an angle how do I calculate the effective g, is it the Sin?

    So let me explain the problem then maybe someone can tell me how to solve it, or in fact is there a problem to be solved?

    If I am in a car travelling horizontally and accelerating at a rate of 1g what is the g I am subjected to?
    __
    Is it 2g or / 2 = 1.414

    Two, if I am travelling in a car up an incline of 45 degrees at a constant 30mph what is the g, and how do I calculate it please…

    Many thanks IMK
    PS This is not a home work question but a problem I am trying to figure out with very rusty physics.
     
  6. Nov 17, 2007 #5

    Doc Al

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    If you are moving at constant velocity the net force on you is zero. That means something is exactly supporting your weight and that your apparent weight just equals your actual weight (mg). Your speed or direction does not affect your weight.

    The fact that you are accelerating makes all the difference! The supporting force of the car must do two things: (1) support your weight with an upward vertical force equal to mg, and (2) push you forward with a force equal to mg to accelerate you. The total force of the car on you (and thus your apparent weight) is [itex]\sqrt{2} mg[/itex], thus you are subjected to [itex]\sqrt{2}g[/itex].


    You got rid of the acceleration! So your apparent weight is back to normal = 1 g.
     
  7. Nov 17, 2007 #6

    IMK

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    Doc AL, many thanks for your reply
    Sorry to bother again, but are you sure that if I am travelling vertically at a constant speed then I will only feel 1g.
    Again many thanks IMK
     
  8. Nov 17, 2007 #7

    Doc Al

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    Of course. But don't take my word for it, figure it out! What one feels is the force of whatever is supporting you--that's your apparent weight and what gives you the feeling of gs. Per Newton's 2nd law, the net force on a body equals "ma". Thus, if something is not accelerating, the net force on it is zero.

    The typical illustration of apparent weight is what you feel when on an elevator. If the elevator is moving at a constant speed, the net force is zero. So the upward force you feel against your feet must exactly equal the force of gravity. The fact that you are moving up or down doesn't matter. If the elevator were moving at some angle rather than straight up, it still wouldn't matter.

    But if the elevator is accelerating upwards, there must be a net upward force. If we represent the force of the elevator pushing against your feet as "N" (for normal force), then:
    The net force on you = +N - mg
    Newton tells us: +N - mg = ma
    Thus: N = ma + mg = m(g + a)
    Which means you feel a "g-force" of "g + a".
     
  9. Nov 17, 2007 #8

    IMK

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    Doc Al,Many thanks and well I have to change the way of thinking about what I am trying to do..
    Again many thanks for your help IMK
     
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