A student stands on a bathroom scale in an elevator at rest on the 64th floor of

  1. 1. The problem statement, all variables and given/known data
    A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 828 N.
    (a) As the elevator moves up, the scale reading increases to 919 N. Find the acceleration of the elevator.
    (b) As the elevator approaches the 74th floor, the scale reading drops to 782 N. What is the acceleration of the elevator?
    (c) Using your results from parts a and b, select which change in velocity, starting or stopping, takes the longer time. Explain.


    2. Relevant equations

    f=ma

    3. The attempt at a solution

    I've done all of these problems twice and I can't figure out the solution. I thought you would just find the average acceleration but I guess not.
     
  2. jcsd
  3. Doc Al

    Staff: Mentor

    Re: A student stands on a bathroom scale in an elevator at rest on the 64th floor of

    Start by identifying the forces acting on the student, then apply Newton's 2nd law: ƩF = ma.
     
  4. Re: A student stands on a bathroom scale in an elevator at rest on the 64th floor of

    I know that the scale is exerting a force but I'm not sure how much. Also, would the elevator be exerting a force? I also know that gravity is a force but I think that's it.
     
  5. Doc Al

    Staff: Mentor

    Re: A student stands on a bathroom scale in an elevator at rest on the 64th floor of

    Right, the scale exerts an upward force. The force it exerts is the reading on the scale.
    No, not directly. (It exerts its force by pushing on the scale. All we care about are forces on the person, and the scale is the only thing touching the person.)
    Good. Only two forces act. Set up an equation from Newton's 2nd law, then you can solve for the acceleration.

    Can you figure out the person's mass?
     
  6. Re: A student stands on a bathroom scale in an elevator at rest on the 64th floor of

    Ok so that would get you 828 N/ 9.8 m/s/s = 84.4898. Is that in kg? So that's the mass of the person so then do 919 N / 84.4898 = 10.88 which is your acceleration?

    And then for the last part it would be 728/84.4898 = -8.62 the negative because it's slowing down. Negative acceleration. Correct?
     
    Last edited: Oct 2, 2012
  7. HallsofIvy

    HallsofIvy 40,541
    Staff Emeritus
    Science Advisor

    Re: A student stands on a bathroom scale in an elevator at rest on the 64th floor of

    I think it would be easier to use the difference in force 919- 828= 91 N. Since The 828 N was the weight of the person standing still, the 91 N must be the result of the upward acceleration of the elevator. NOW use "f= ma". What is m, the mass of the person?
     
  8. Re: A student stands on a bathroom scale in an elevator at rest on the 64th floor of

    Ok so acceleration would be 9.29 m/s/s and then for the second part it would be -4.70! I get it thank you :D
     
  9. Doc Al

    Staff: Mentor

    Re: A student stands on a bathroom scale in an elevator at rest on the 64th floor of

    You're off by a factor of 10. I suggest redoing them more carefully (especially the second part).

    Don't skip steps. Write out the equation:
    ƩF = ma
    scale force - weight = ma
     
  10. Elevator Standing Still: Normal Force (reading on scale)= Weight of person = mg
    The Normal force required is enough to balance the downward force created by the mass (person) wanting to accelerate downward at 9.8 m/s/s. (the scale/floor of elevator prevents them from plummeting to their death)

    Elevator accelerating upward: Normal force(reading on scale)= mg (for the same reasons as stated above) + ma. with "a" being upward acceleration of the mass (person) caused by the scale/elevator floor. (F=ma) since the force required to cause the upward acceleration opposes gravity, (just like the force required to prevent the person from plummeting to their death)

    Elevator accelerating downward: same theory acceleration upward but since acceleration is in the same direction as gravity, it reduces the Normal Force required to keep the mass from plummeting. (you can subtract a positive value or add a negative value but either way, the Normal Force (reading on the scale) is reduced.
     
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