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What Fraction Of Her True Weight Was Indicated?

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data
    Julie tightened herself fastened a set of bathroom spring scales to a platform and went down a 30 degree slope. No friction was present and the platform supporting the scales was HORIZONTAL. what fraction of her true weight was indicated by the scales?



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    3. The attempt at a solution

    so what i did was just assume a random mass, in this case 50kg so that means around 500 Newtons.
    Therefore, Fnormal would be 500 cosine 30 = 433 Newtons
    than all i did was take 433 and divide it by 500 which gives me 0.866. So is it right to say that 87% of her true weight will be shown on the scale?
     
  2. jcsd
  3. Mar 27, 2012 #2

    PhanthomJay

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    That is not correct. The normal force you have calculated is the force of the plane on the wheels , acting perpendicular to the incline. And that would be the scale force indication if the scale was also parallel to the inclined slope. But it is given that the scale platform is horizontal. You've got to look at the forces acting on the person in the vertical and horizontal directions. Is there acceleration in the vertical direction?
     
  4. Mar 27, 2012 #3
    Well there is no acceleration given but cant you say that 9.81m/s^2 is the acceleration due to gravity? and if you do the forces acting in the vertical direction wouldnt it just be 500N up and than fg = 500 N down?
     
  5. Mar 27, 2012 #4

    PhanthomJay

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    well yes, but the acceleration due to gravity is not the vertical acceleration of the person and scale and platform
    No, that would be true only if the net force in the vertical direction was 0 (no acceleration in the vertical direction). But there is acceleration in the vertical direction, because the person is moving vertically downward at increasing velocity due to a net force downward in that vertical direction. The person is also accelerating horizontally, because he/she is also moving horizontally at increasing speed. Note further that the person's resultant acceleration is parallel to the slope, so you might want to find the acceleration of the platform and person and scale down the plane in the usual manner you are accustomed to for forces acting parallel to the plane. Then find the vertical component of that acceleration and use newton 2 in the vertical direction.
     
  6. Mar 27, 2012 #5
    So the acceleration parallel to the surface is just mg sin theta correct? So, if assuming a random mass of 10kg than acceleration parallel to the surface = 10(9.81) sin 30 = 49.05m/s^2 - isnt that too fast? or should i not assume a weight at all? And how would i find acceleration vertically?, is it (assuming mass of 10kg) mgcos30?
     
  7. Mar 27, 2012 #6

    PhanthomJay

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    The force parallel to the incline is mg sin theta. The acceleration parallel to the incline is ___?___
    correct this once you determine the correct value for the acceleration down the incline (call it ap)
    It's OK to assume it, or just calling it 'm' is better
    No, the vertical component of the acceleration, ay, comes from the trig... ay/ap = ___?___ (it is not the cos function, draw a sketch)...
     
  8. Mar 27, 2012 #7
    Oh ya my bad mg sin theta = the force so in order to find acceleration it would be
    mg sin theta/m. So you can actually just cancel out mass i guess right? so a = 49.05/10 = 4.905m/s^2.
    so...

    ap = 4.905m/s^2
    is ay than just Fnet/m? Or is it just 9.81m/s^2..im still not too sure about the vertical acceleration.....
     
  9. Mar 28, 2012 #8
    Wait i think i just got it. With the angle 30 degrees the ay = sin30 = ay/ap = sin30 = ay/4.905m/s^2

    that solves to be 2.4525m/s^2 = ay

    I can also solve for ax which would equal cos30 = ax/ap = ax /4.905 = 4.24m/s^2

    So acceleration in the horizontal equals 4.24m/s^2 and the vertical acceleration = 2.4525m/s^2that is parallel to the surface = 4.905m/s^2

    I am honestly clueless however to find the fraction of her true weight
     
    Last edited: Mar 28, 2012
  10. Mar 28, 2012 #9
    Wait to solve for the fraction of the true weight, would i take acceleration in the y (2.4525m/s^2) and subtract it from gravity so 9.81 - 2.4525 = 7.3575m/s^2 and than multiply that by 10 which gives me 72.2 Newtons. And than would i take 9.81 x 9.81 = 96.2361 Newtons and than divide 72.2/96.2361 = roughly 3/4 her true weight?
     
  11. Mar 28, 2012 #10
    Wait would i do the 9.81 x 9.81 part??
     
  12. Mar 28, 2012 #11
    I think you have to compare this with a girl inside an elevator.
    If accelerating upward the weight increases, N-mg=a
    In your case find downward acceleration and you will get N, the weight.
     
  13. Mar 28, 2012 #12
    Ya I found downward acceleration which Is 2.45m/s^2 how from here can I get the weight?
     
  14. Mar 28, 2012 #13
    You should not use random value. Just use mg for the weight.
    It just a ratio of apparent weight over normal weight.

    What is the apparent weight in term of angle.
    You can predict when the angle increase, the acceleration increase too.
     
  15. Mar 28, 2012 #14
    Ok so how would I take my downward component of 2.45m/s^2 and use itto find weight?
     
  16. Mar 28, 2012 #15
    Its best to substitute the value until the last(if possible).
    You should find what is the acceleration in term of the angle 30°.

    Then apparent weight Mg-N=ma , N=Mg-ma

    The ratio will be Mg-Ma/Mg
     
  17. Mar 28, 2012 #16
    wait, does my vertical acceleration look correct? if correct would I just use your equation n-mg= a? But how do I work the angle in??
     
  18. Mar 28, 2012 #17
    ya I found the acceleration in terms of the, angle butshouldnt I use the vertical acceleration because the scale on the surface is horizontal?
     
  19. Mar 28, 2012 #18
    You can try. Once you get the answer then check what happen if the angle is 0 and then if the angel is 90°. Thus the value makes sense?
     
  20. Mar 28, 2012 #19

    PhanthomJay

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    Yes, her downward acceleration is 2.45 m/s^2, very good. Her weight is just mg or 98N if you assume m = 10 kg. Her apparent weight is the scale reading, or the normal force acting perpendicular to the scale. You actually had the correct answer in your post #9, (but your input values were slightly off.... ). Using newton 2 in the vert direction as per azizlw post #11, you get the answer for N. If you had not assumed a mass and instead used her mass as m, you get the normal force directly as a fraction of her weight.
     
  21. Mar 28, 2012 #20
    What is wrong with my input values?
     
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