1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Elevator kinematic

  1. May 18, 2003 #1
    Hi,

    here is the question:
    there is a elevator, an man is in it, via a pulley he can raise the elevator.

    zzzzz/o\
    zzzzz|zz\
    zzzzz|zz|
    ---zz|------
    |zzzz|zzzzz|
    |zzzz|zzzzz|
    |zzz\o/zzzz|
    |zzzz|zzzzz|
    |zzz/z\zzzz|
    ------------
    This is my picture so can't do better
    z are empty space as this forums gets rid of them
    there is a man in the middle of the box, pulling a pulley.
    the elevator is 30kg the man is 70kg,
    this man exert a force of 300N to rise himself, we need to find the acceleration.
    the answer is 5m/s^2 but i can't find why.
    What i do is that

    the mass of the whole system is 100kg (man + elevator)
    the tension in the string is 300N, and using the formula
    Mg-T=ma
    where g = 10
    700-300=100a
    4m/s^2
    any feed back is appreciated
     
    Last edited by a moderator: May 18, 2003
  2. jcsd
  3. May 23, 2003 #2

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I don't have a calculator handy, but can you check to see if the error comes from the approximation (g=10) you used? Try again with g=9.8 and see if it doesn't come out right.
     
  4. May 24, 2003 #3
    the acceleration is 10g.
    But i can't figure out how to do it
    I want to do
    T-30g=30a
    100g-T= 100 a (100 = 70+ 30)
    if i add them together i get
    700 = 130 a
    a= 5.4m/s

    But this is how close i get to the answer!!
    and i'm sure i am wrong because the T = 300
    so g = 10
    in the first equation
    T-30g=30a
    300-30*10=30a
    0=30a
    so i can't go nowhere here.
     
  5. May 24, 2003 #4
    Hey, I posted an answer to this a few days ago (OK - it was slightly "camouflaged" with a few extra z's, but all you had to do was search/replace the z's with spaces) but that entire post somehow got erased.

    Down with censorship!!

    Anyway, am I getting the picture right? The guy is standing in an elevator, there is a rope attached to the top of the elevator, the rope goes over a pulley attached to the top of the elevator shaft, and the guy is pulling on the other end of the rope with 300 N force?

    There's no way that elevator is going up!!

    If the guy pulls on the rope with 300 N force, the rope pulls upward on him with 300 N force. And, assuming the tension everywhere in the rope is 300 N, the rope also pulls upward on the top of the elevator with a force of 300 N. So, total upward force = 600 N.

    The downward force is (30 kg + 70 kg) x 10 m/s2 = 1000 N.

    Net force (downward +) = 1000 - 600 = 400 N.

    a = f/m = 400/100 = 4 m/s2 (downward)

    If you use g=9.8, the acceleration is 3.8 m/s2 (still downward)
     
  6. May 24, 2003 #5
    Hi,

    now the answer is 5m/s upwards,
    because the man insinde the elevator is pully himself up with the rope.

    imagine: there is a box (elevator with a hole just to let through a rope).
    now this rope passes through a pulley and is attach to the top of the elevator. inside the elevator a man is hanging on the rope, and is pulley, exerting a force of the 300N so both the man and the elevator is going upwards, now we need to find this acceleration upwards.
     
  7. May 24, 2003 #6
    Impossible!

    Forget about formulas & think about this logically. The man is not even exerting enough force to lift himself, without an elevator. How could he possibly be lifting himself AND the elevator?

    To accelerate upwards at 5 m/s2, he would have to pull on the rope with a force of 750 N. (The tension would have to be 500 N just to stand still.)
     
  8. May 25, 2003 #7

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I thought you were just banging on the keyboard!
     
  9. May 28, 2003 #8
    are you guys forgetting the force multiplication from the pulley?
    BTW with first method I recieve:
    mg - T = ma
    a = g-T/m = (10 m/s2 - 300N )/100kg) = ( (10)-(3) )=7 m/s2
    (T/m = (car+man kg)*10m/s2)
     
    Last edited: May 28, 2003
  10. May 28, 2003 #9
    HUH???
    You are forgetting the force multiplication from the pulley.
    Also, you wrote
    a = g-T/m
    but then you did
    a = (g-T)/m

    (Doesn't matter, though, since both are wrong.)

    You have to do
    mg - 2T = ma
    a = (mg - 2T)/m = ((70+30)*10 - 2*300)/(70+30)
    a = 4
    and remember that if g = 10 m/s2 downward, then
    a = 4 means 4 m/s2 downward
     
  11. May 29, 2003 #10
    Aha!
    very simple:
    ∑Fx = 0
    ∑Fy = T - m1g = m1a
    ∑Fy = m2g - T = m2a
    add 2nd equation to 1st:
    -m1g + m2g = m1a + m2a
    --> a=((m2-m1)/(m1+m2))g = ((300 - 100)kg/(300+100)kg)9.8m/s2=4.9 m/s2

    EDIT
    with g = 10 m/s2 a = 5 m/s2, just like thread author reported.
     
    Last edited: May 29, 2003
  12. May 29, 2003 #11
    [zz)]

    I'da thought the guy would lose weight from all that exercise. Yet somehow he seems to have grown from 70 kilos to 300, and even stranger, the elevator put on some weight too, increasing from 30 kg. to 100. Where'd you get those numbers?

    Going back to the original numbers:

    M1 = 30 kg. (elevator)
    M2 = 70 kg. (man)
    T = 300N

    With those numbers, using your method,
    T - m1g = m1a
    m2g - T = m2a
    a = g*(m2-m1)/(m2+m1)
    a = 10*(70-30)/(70+30)
    a = 4 m/s2
    This might look like it agrees with my answer, but actually it's completely wrong. Notice that the way you set up the equations, up is the positive direction (since in your ∑Fy you're subtracting m1g instead of adding). So this answer implies that a is 4 m/s upward.

    Now let's put this result back into your original equations. Now we have
    T - m1g = m1a
    300 - 30*10 = 30*4 => 0 = 120
    and
    m2g - T = m2a
    70*10 - 300 = 70*4 => 400 = 280

    What went wrong? Doesn't it strike you as strange that you can completely eliminate T from the calculation? Meaning that the acceleration is the same no matter how hard the man pulls on the rope?

    You can set it up that way only if the man is not standing in the elevator. Then the man accelerates down and the elevator accelerates up, and T is determined by their relative weights, not by how hard the man wants to pull.

    But in this problem, the man is standing in the elevator. You omitted from your equations the force that the elevator exerts on the man, and the force the man exerts on the elevator. These are obviously equal and opposite, and we don't know the magnitude of that force, so call it R.

    Now the equations are:
    T - m1g - R = m1a for the elevator
    T - m2g + R = m2a for the man
    Add them together. Now it makes sense that R disappears, since it is an internal force within the man/elevator system:

    2T - m1g - m2g = m1a + m2a
    a = ((2T - (m1+m2)g)/(m1+m2) looks familiar?
    a = (600 - (30+70)*10)/(30+70) = -4 m/sec2 as previously claimed. We set up the equations with up being positive, so this acceleration is downward.

    Now you can plug that back into either of the equations and find that
    R = T - m1a - m1g = 300 -(-120) - 300 = 120N
    R = m2a + m2g - T = -280 +700 - 300 = 120N
     
    Last edited: May 29, 2003
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Elevator kinematic
  1. Elevator ? (Replies: 4)

  2. Find the elevation (Replies: 12)

  3. Elevator Problem (Replies: 5)

Loading...