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Basic Apparent Weight Question

  1. Feb 2, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    A 60 kg person rides in a 20 kg cart
    moving at 13 m/s at the bottom of a valley that is in the shape of an arc of a circle
    with a radius of 36 m. What is the apparent weight of the person as the cart passes
    the bottom of the valley?

    2. Relevant equations
    F_centripital = m*v^2/r
    F = ma

    3. The attempt at a solution

    So this seems a little too easy to be correct.
    Apparent weight = m(a+g)=m(v^2/r+g)
    =60(13^2/36+9.80)
    =869.67 N

    The thing I am wondering about is the sign on gravity according to it being in the bottom of a valley.
    Is my work correct?
     
    Last edited: Feb 2, 2015
  2. jcsd
  3. Feb 2, 2015 #2
    Question, where is the centripetal force directed and where is the gravitational force directed?
     
  4. Feb 2, 2015 #3

    RJLiberator

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    The centripital force is directed upwards and the gravitation force is directed downwards.

    This would imply that the correct equation would be: m(a-g) = m(v^2/r-g) ?
     
  5. Feb 2, 2015 #4
    Yes! or (g-a) depending on how you chose your coordinate system.
    From this, we will have a smaller magnitude of apparent weight. This creates the feeling of lightness when you are going fast down a hill in your car, or a situation similar.
     
  6. Feb 2, 2015 #5

    RJLiberator

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    60(13^2/36-9.80)
    nets me an answer of -306.33N
    The negative apparent weight implies that the body is being accelerated downward with an acceleration greater than that exerted by the Earth.
    Is this a correct understanding?
     
  7. Feb 2, 2015 #6
    Well, if you considered just the weight of the person stationary, 60 x -9.8 = -588 N. You computed -306.33 N, he feels less of a pull then he would normally by gravity, i.e. feeling lighter.

    The forces oppose each other so their sum cannot have a magnitude greater than the individual forces.
     
  8. Feb 2, 2015 #7

    RJLiberator

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    Ahhhh, now that makes sense to me.

    So regularly, he is being pulled at -588 N. In my original answer of 806 he would have been feeling heavier which wouldn't make sense.
    Since we updated it to be -306.33 N, we see that he is feeling relatively lighter.

    Great. Thank you kindly for your help.
     
  9. Feb 2, 2015 #8
    No problem, happy to help!
     
  10. Feb 2, 2015 #9

    RJLiberator

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    Two more things regarding this thread:
    1) Why was the 20kg of the cart neglected throughout this entire problem? Was this thrown in just to throw me off?
    2) The force vectors in a full-body diagram would be: Force of gravity pointing downwards which would be equal to the centripetal force which points upwards towards the center of the "circle" and then a velocity vector pointing to the right perpendicular and a friction vector pointing the opposite way, but smaller than the velocity vector.
    Correct?
     
  11. Feb 2, 2015 #10
    1) Looking at the problem again, we should have done 60 + 20 for the mass the entire time. So readjust your calculations with 60+20 as the mass. I went off the number you used originally (60) in your attempted answer so I did not notice that there was the 20 kg cart.
    EDIT: nevermind, it asks for the apparent weight of the person and not the apparent weight of the cart + person system.

    2) If you are just drawing the force vectors, do not include the velocity vector since that is not a force. As with friction, that is a force so it would be appropriate to draw that IF friction is applied to the problem (which it doesn't specifically mention).
     
  12. Feb 10, 2015 #11

    RJLiberator

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    I'm looking over this problem again, trying to understand everything:

    A) So the stationary person feels a -784N force? How can someone feel a negative force?

    And then B) When he is at the bottom of the cliff we calculated a -408.4N force and since this is more in the positive direction, this is a heavier force?

    Hm
     
  13. Feb 10, 2015 #12
    Negative meaning opposite the direction you chose to be positive (ex up is positive so down is negative, or down is positive so up is negative)

    For the second paft, the magnitude of the resulting force is less (400 < 700) so he is less pulled down then he would normally be.
     
  14. Feb 10, 2015 #13

    RJLiberator

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    Ah, the absolute value will dictate the force. The orientation of my coordinate system dictates whether it will be negative or positive. Thank you kindly.
     
  15. Feb 10, 2015 #14
    No problem, happy to clarify!
     
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