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Problem concerning Pilot find true weights

  1. Oct 27, 2011 #1
    1. The problem statement, all variables and given/known data
    The Pilot of an airplane executes a loop-the-loop manuever in a vertical circle. The airplane's speed is 300 mi/h at the top of the loop and 450 mi/h at the bottom. The radius of the circle is 1200 ft. Use American Engineering units in this problem.

    A) The pilot's apparent weight is equal to the magnitude of the force exeted by the seat on his body. What is the pilot's weight at the circle's highest point if his true weight is 160 lb?

    B) What is the pilot's apparent weight at the circle's lowest point?

    2. Relevant equations
    I know I have to find the some of forces and solve for his weight.


    3. The attempt at a solution

    I know the first step is to draw a FBD and label the forces acting on the pilot.
    At the top of the look, with the pilot travelling 300 mi/hr the forces acting are...

    Gravitational Force = mg = 32 ft/s^2
    w/Pilot's weight = 32 x 160lb = - 5120
    Normal Force of seat = Fn = Fg = 5120 ??

    I know I have to factor in the velocity but im not sure what the equation is that i have to use...
    Please help thanks
     
  2. jcsd
  3. Oct 27, 2011 #2

    PeterO

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    What are the base units in the "American Engineering Units"

    SI units have length as metres, mass as kg, Force as Newtons, time as seconds.

    What do "American Engineers" use?
     
  4. Oct 27, 2011 #3
    I believe my teacher is talking about using units such as lbs and feet.
     
  5. Oct 27, 2011 #4
    You know that if you are free falling that your weight is 0. it is due to your acceleration downwards being equal to that of gravity. If you are sitting on the ground, your weight is equal to the normal force of the ground pushing back on you.

    to relate this to your question, you know that regardless of where you are on the circle that the acceleration is towards the centre. If you're at the top of the circle, you have a downards acceleration equal to v^2 / r.
     
  6. Oct 27, 2011 #5
    So his weight will be the acceleration downwards?
    v = velocity?
    r = radius ?
     
  7. Oct 27, 2011 #6
    yes

    His weight is a bit tricky to think about in terms of this example. Try thinking about it as if you are in an elevator (since we are only really considering acceleration in the vertical direction)

    If you accelerate upwards in an elevator, you feel more weight, for the following reason.

    When the elevator accelerates upwards, the bottom of the elevator pushes you upwards with a certain force. But while it is pushing you upwards, gravity is also pushing you downwards. this gravity in turn causes a normal force once again from the floor of the elevator. This normal force comes from the sum of gravity and the acceleration.

    does that make any sense to you?

    In the case that the elevator accelerates downwards, you feel a little weightlessness at the peak of acceleration. This is because you are closer to being in a free fall. since the elevator accelerates downwards, the acceleration due to gravity doesn't harvest in the form of weight, it harvests in the form of speed.
     
  8. Oct 27, 2011 #7
    This makes alot of sense, and living in NYC I can definately relate to what your saying with the elevators. So my answer is going to be the Normal Force...
    Fn = gravity + acceleration
    = (-mg) + (v^2/r)
    **My next question is this...in this equation gravity will be negative because it is going against the Normal force right? but if this is the case should the acceleration also be negative? because it is in the downward direction as well?**
    Also in discussing this problem with my friend he suggested a similar equation:
    Fn= -mg +mv^2/r
    He added the mass of the man in the acceleration aspect of the problem, is that even needed if like you said he is at free fall therefore his weight is 0?
     
  9. Oct 27, 2011 #8
    That equation seems right but I don't quite understand your last question there. Weight is a unit of force which is mass * acceleration. On the moon you weigh less because there is less gravity.

    your problem is tricky in terms of signs. You have to be careful about which way the normal force is being applied, and how that interacts with gravity. It might be useful to do this whole problem in space first (no gravity), and then add in that "-mg" factor and see if it coincides with your intuition. If it doesn't you've flubbed a sign.
     
  10. Oct 27, 2011 #9

    PeterO

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    Yes, but what sort of pounds.

    In England and Australia, lbs are a unit of mass, like kilograms.

    In some American situations lbs are a unit of force, like Newtons.

    When they say the man weighs 160lb, that means in England he has a mass of 160lb, but in America he has a weight of 160 lb.

    Also, this question specifically relates to apparent weight - which is how strongly the environment pushes - contact force.

    AT the top of the flight, speed = 300 mph = 440 ft/sec

    Using v2/R that gives a centripetal acceleration of 161.33

    Gravity would provide 32 of that. The seat must supply the other 129.33

    The weight force of 160lb would provide an acceleration of 32 [if this pilot jumped off a roof] so a force slightly more than 4 times is needed for this acceleration of 129.33.

    So I think his apparent weight would be a little over 640 lb.

    AT the bottom, the acceleration is far greater, and upwards.
    The seat will have to push to balance g, and provide all the acceleration.

    lets suppose the acceleration was 320 [makes the calculations easy].
    The seat would have to push up with 160lb, to overcome gravity, and a further 1600lb to provide this acceleration.
    His apparent weight would by 1760 lb.

    Now you can calculate the real acceleration, and thus get the actual apparent weight.
    Even the top of the loop has to be calculated accurately.
     
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