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True vs. Apparent weight question

  1. Apr 20, 2016 #1
    1. The problem statement, all variables and given/known data
    An astronaut in a rocket has an apparent weight of 1.35x103N [down]. If the acceleration of a rocket is 14.7 m/s2 [up] near Earth's surface, what is the astronaut's true weight? The acceleration due to gravity on Earth's surface is about 9.81 m/s2 [down].
    Mass of Earth is 5.97x1024, not sure if that's needed or not.
    2. Relevant equations
    Fnet = ma
    Fg = Gm1m2 / r2
    g = Gm(source) / r2
    g = Fg / m(test)
    3. The attempt at a solution
    I'm confused about what I should be solving for (I think it should be Fg), if the apparent weight is the normal force shouldn't it be directed up? I'm also confused about which acceleration to use and where, also which formula to use. So I'm pretty lost! Any help would be greatly appreciated :)
     
  2. jcsd
  3. Apr 20, 2016 #2

    PhanthomJay

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    Yes the normal force is directed up on the astronaut and is equal to the so-called apparent weight. Just use Newton's 2nd law and your first and last equation after first determining the other force besides the normal force acting on the astronaut. No need to use the middle 2 equations since g is already given.
     
  4. Apr 20, 2016 #3
    Have you drawn a free body diagram, or do you think you have advanced beyond the point where you need to use free body diagrams?
     
  5. Apr 20, 2016 #4
    I'm still a bit confused, I solved for the mass using Fg = (mass)(gravity) :
    1.35x103N = (mass)(9.81m/s2)
    mass = 137.614 kg

    But then after I plugged it into my formula Fn - Fg = ma I got a true weight of -672 N. I'm not sure where I went wrong.
     
  6. Apr 20, 2016 #5
    I have drawn the free body diagram, and I have come up with the equation Fn - Fg = ma. However when I plug in the mass that I found (if that's even the right mass) I get a negative answer.
     
  7. Apr 20, 2016 #6
    How much is the system accelerating by?
     
  8. Apr 20, 2016 #7

    PhanthomJay

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    You have mistakenly confused apparent weight with true weight. Fg is true weight, Fn is apparent weight. Your equation is correct, but you need to make the proper substitutions. The normal force is not equal to mg.
     
  9. Apr 20, 2016 #8
    14.7 m/s2 up.
     
  10. Apr 20, 2016 #9
    Oh I see! I know why that is wrong, but I'm still not sure about how to find the mass of the astronaut correctly. I think I need that first to answer the question.
     
  11. Apr 20, 2016 #10
    Good. Now plug in ##F_g=mg## and solve for ##F_n##
     
  12. Apr 20, 2016 #11

    PhanthomJay

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    Yes that is correct. F_n is given. F_g is mg. Solve for m using your equation that you determined from your free body diagram and Newton 2, substituting in the given value for a. Then solve for F_g.
     
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