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Elevator problem with safety rails, spring, and a snapping cable->

  1. Oct 22, 2004 #1
    Elevator problem with safety rails, spring, and a snapping cable--->

    The cable of an 1800 kg elevator snaps while it is 3.7 meters above a spring with constant of k = 0.15 MN/m. Also, a safety guide rail device provides a 4.4 kN frictional force against the fall for the duration of the fall.

    a) What is speed of cab just before hitting the spring?

    b) What is the maximum distance x that the spring is compressed (frictional force still applies during the compression of the spring)?

    c) How far will the cab bounce back up the shaft?

    d) What is the total distance the cab will move before coming back to rest?

    Help, I have been trying to figure this out for 3 hours!
     
  2. jcsd
  3. Oct 23, 2004 #2

    Pyrrhus

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    what have you done?
     
  4. Oct 24, 2004 #3
    I am not really sure if this is correct for the first part of the problem (a):

    I used the equation [itex]v^2 = v^2_0 + 2a(x-x_0)[/itex] and used [itex]0[/itex] for initial velocity and [itex]9.8 m/s^2[/itex] for a and [itex]3.7 m[/itex] for [itex]x-x_0[/itex] (also displacement d)

    Substituting, I get: [itex]v^2 = 19.6 m / s^2 * 3.7 m = 72.52 m^2 / s^2[/itex] and then [itex]v = 8.52 m / s[/itex], when the cab hits the spring, right? Or am I going about this all wrong?

    I think that I am forgetting the frictional force acting on the cab in a constant manner opposing it's fall, it says that this force is 4400 N, how would I "subtract" thhis from the 8.52 m / s?
     
    Last edited: Oct 24, 2004
  5. Oct 24, 2004 #4

    Doc Al

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    Nothing wrong with using kinematics to find the speed. But you'd better use the correct acceleration. The acceleration would be 9.8 m/s^2 if gravity were the only force acting on the cab. But friction acts. So find the net force on the cab, then use Newton's 2nd law to find the acceleration.
     
  6. Oct 24, 2004 #5
    How do I find the net force then?
     
  7. Oct 24, 2004 #6

    Doc Al

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    You tell me. What forces act on the elevator cab?
     
  8. Oct 24, 2004 #7
    The force of gravity, [itex]mg[/itex], which is [tex]1800 kg * 9.8 m / s^2 = 17640 N[/tex]. Then the force of the friction, [tex]f_k[/tex], which is [tex]ma[/tex], which I know [tex]mass = 1800 kg[/tex], but I don't know acceleration.
     
  9. Oct 24, 2004 #8

    Doc Al

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    Good.
    The force of friction is given. It is not equal to ma: Newton's 2nd law says that the net force on an object equals ma.

    Here's how to think of it. There are two forces acting on the cab:
    (1) the weight, acting down (you calculated this)
    (2) the friction, acting up (this is given)

    Find the net force by adding these two forces (direction counts--forces are vectors). Then figure out the acceleration.
     
  10. Oct 24, 2004 #9
    Is the net force on the cab then [tex]17640 N - 4400 N = 13240 N[/tex]?
     
  11. Oct 24, 2004 #10

    Doc Al

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    Right! That's the net force, which acts down. Use that to find the acceleration.
     
  12. Oct 24, 2004 #11
    The net acceleration of the system is then [tex]13240 N / 1800 kg = 7.35 m / s^2[/tex]? How do I use this to find the velocity after falling [tex]3.7 m[/tex]?

    Does the velocity come from the same equation [itex]v^2 = v^2_0 + 2a(x-x_0)[/itex]? If so, I get [tex]v^2 = (0)^2 + 2(7.35 m / s^2)(3.7 m)[/tex]
    And then: [tex]v = \sqrt{54.39} = 7.37 m / s[/tex] < ---- Is this correct?
     
    Last edited: Oct 24, 2004
  13. Oct 24, 2004 #12

    Doc Al

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    Looks good to me.
     
  14. Oct 24, 2004 #13
    About part b), I have this so far:

    [tex]0 - \frac{1}{2} m v^2_i = m g (d) - \frac{1}{2} k (d)^2[/tex]
    [tex]-\frac{1}{2} (1800 kg)(7.4 m / s)^2 = (1800 kg)(9.8 m / s^2) (x_{max}) - \frac{1}{2}(150000 N / m)(x_{max})^2[/tex]

    Eventually I calculate that the [tex]x_{max} = -0.70 m[/tex] or [tex]x_{max} = 0.94 m[/tex], in which case I would choose [tex]0.94 m[/tex], right? Does this all look correct or am I way off again?
     
    Last edited: Oct 24, 2004
  15. Oct 24, 2004 #14

    Doc Al

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    energy lost to friction

    You forgot to account for the energy lost to friction. (Mechanical energy gets transformed to heat.) The initial mechanical energy is KE + Gravitational PE; the final is Spring PE. The difference is the work done against friction.
     
  16. Oct 25, 2004 #15
    How would I go about accounting for this lost energy due to heat? What formulas would I use?
     
  17. Oct 25, 2004 #16
    the energy due to friction would be the magnitude of the frictional forces times the distance it moved along (parallel to the motion)
     
  18. Oct 25, 2004 #17
    Would it then be:

    [tex]-\frac{1}{2} (1800 kg)(7.4 m / s)^2 = (1800 kg)(9.8 m / s^2) (x_{max}) - \frac{1}{2}(150000 N / m)(x_{max})^2 - (4400 N)(3.7 m)[/tex]?

    Then I get: [tex]x_{max} = 0.79 m[/tex], does that sound right?
     
    Last edited: Oct 25, 2004
  19. Oct 25, 2004 #18

    Doc Al

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    Why are you using 3.7m? That was the distance used in part a; it's not relevant here. For part b, over what distance does friction act?
     
  20. Oct 25, 2004 #19
    3.7 m is the distance the elevator cab falls before reaching the spring, then the spring compresses an additional [tex]x_{max}[/tex] m and then the total distance would be [tex](3.7 + x_{max}) m[/tex]?

    Is this right:

    [tex]-\frac{1}{2} (1800 kg)(7.4 m / s)^2 = (1800 kg)(9.8 m / s^2) (x_{max}) - \frac{1}{2}(150000 N / m)(x_{max})^2 - (4400 N)(3.7 + x_{max})[/tex]

    [tex]x = 0.83 m[/tex]? Or am I eternally doomed to be a physics retard?
     
    Last edited: Oct 25, 2004
  21. Oct 25, 2004 #20

    Doc Al

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    In your analysis for part b you are comparing energy at two points:
    (1) where the cab first reaches the spring (you calculated its speed at that point)
    (2) where the spring is maximally compressed

    In moving from point 1 to point 2, the cab moves a distance of [itex]x_{max}[/itex]. The 3.7 m that the cab moved prior to reaching point 1 is no longer relevant (at least in this analysis).
     
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