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Need a check on a elevator going up problem

  1. Feb 16, 2012 #1


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    Just want to see if I did everything right. I'm notorious for simple mistakes.
    1. The problem statement, all variables and given/known data
    A 100kg man concerned about his weight decides to weigh himself in an elevator. He stand on a bathroom scale in an elevator that is moving upward at 3.0 m/s. As the elevator reaches his floor, it slows to a stop.

    If the elevator slows to a stop over a distance of 2.0m, what is the reading on the bathroom scale?

    2. Relevant equations
    I used
    F = ma
    v2 = v1 + a12(t2 - t1)
    r2 = r1 + v1(t2 - t1) + (1/2)(a12)((t2-t1)^2)

    3. The attempt at a solution
    Ok I know that I needed to find the force of the scale. to find that I needed the net force, since I have the force of the gravity on the guy (100*9.8 = 980N)
    But to find that I needed the acceleration.

    thus with v2 = v1 + a12(t2-t1) I got this
    0 = 3 + a12(t2 - 0)
    -3 = a12(t2)
    a12 = -3/t2

    I plugged that into r2 = r1 + v1(t2-t1) + (1/2)(a12)((t2-t1)^2)
    2 = 0 + 3(t2 - 0) + (1/2)(-3/t2)((t2-0)^2)
    2 = 3t2 - (3/2)(1/t2)(t2^2)
    *Canceling I get
    2 = 3t2 - 3t2/2
    4 = 6t2 - 3t2
    3t2 = 4
    t2 = 4/3

    Plugging back into a12 I get
    a12 = -3/(4/3)
    a12 = -9/4

    Thus the net Force = 100(-9/4) = -225N
    Which means Fsacle = -225 - 980 = -1205N
    Making the scale read 1205N since it can only do positive right?
  2. jcsd
  3. Feb 16, 2012 #2


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    Your value for the acceleration is correct, but just FYI, you could have found it much faster by using the equation[tex]v^2 = v_0^2 + 2ad [/tex]where v0 is the initial speed, v is the final speed (at the end of the displacement d), and a is the acceleration. You're given v, v0 and d, and you're NOT given t. So don't choose the kinematics equation that depends on t (an unknown) when you already have equation that depends only on known quantities. Solving for a, you get[tex]a = -\frac{v_0^2}{2d} = -\frac{9.0~\textrm{m}^2/\textrm{s}^2}{4.0~\textrm{m}} = -2.25~\textrm{m}/\textrm{s}^2[/tex]

    As for the rest of the problem: remember that a scale always shows you the normal force between you and it. So that is what you should be trying to compute in this problem: the normal force. Unfortunately, you messed up your force balance equation. That much is obvious for the following reasons:

    - It's true that a scale only shows you the magnitude of the force acting on it. However, in this case, your result for the normal force should have been positive anyway, because the normal force points upward, and you chose upward to be the positive vertical direction.

    - Since the acceleration was negative (downward), this meant that the net force was downward, which means that the normal force must have been pushing up on the person with a force LESS THAN the weight of the person. (To get a net downward force, the upward force from the scale has to be smaller in magnitude than the downward force from gravity). In other words, your final answer should have been less than mg (in magnitude). Yours was greater than mg.

    Let's try writing the force balance equation properly[tex]F_{\textrm{net}} = F_{\textrm{N}} + F_{\textrm{g}} [/tex]where Fg is the gravitational force and FN is the normal force. Now, I'm going to use g = +9.81 m/s2, which means that Fg = -mg, with the negative sign indicating that the force is downward. So the equation becomes: [tex]ma = F_{\textrm{N}} - mg [/tex]You should be able to solve for FN from this equation. Remember that we are using g = +9.81 m/s2, and we found that a = -2.25 m/s2. If you don't get the signs right, you won't get the right answer. You should get a positive answer that is smaller than mg.
  4. Feb 16, 2012 #3


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    Ok Thanks. My professor didn't give us that equation so I'm not sure if we can use it.
  5. Feb 16, 2012 #4


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    The kinematics one? My answer would be that of course you can use it, it is a valid physics equation (i.e. one that is TRUE). However, maybe you have a silly teacher. In that case, you can derive it from the ones that you already know. Start with[tex] v = v_0 + at [/tex]and solve this equation for t, i.e. [tex] t = \frac{v-v_0}{a}[/tex]Then plug this expression for t into the equation[tex]x-x_0 = v_0t + \frac{1}{2}at^2[/tex]Et voila! You will obtain the equation I gave you in my first post (after a bit of algebra). You have successfully eliminated the time variable. This wouldn't save you any work, but at least your teacher would have no cause to complain, because you have then SHOWN it to be true using the physics that he has already taught.

    Of course, a much easier way to derive this equation is to use the work-energy theorem (equate the work done on the object to its change in kinetic energy). But if you haven't covered the concepts of work and energy yet, then I guess that wouldn't be possible.
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