1. Limited time only! Sign up for a free 30min personal 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!

Problem: Braking an elevator

  1. Sep 23, 2016 #1
    1. The problem statement, all variables and given/known data
    A lifting system can be modelled as two blocks of mass m1 and m2 hanging from two pulleys, connected by ropes to a third block of mass M = 100 kg, lying on top of a horizontal surface. There is gravity g = 9.80 m/s2.

    a) Ignoring at first friction, find an expression for the acceleration of the blocks, as a function of the masses and gravity.

    b) Now, we assume that the system starts at rest, and that there is static friction between the block M and the surface it is lying on. How much diffe- rence in mass can m1 and m2 have, for the system to not start moving, if the coefficient of static friction is μs = 1.00? Provide an expression as well as a numerical value.

    c) Next, we assume that the system is moving with an initial speed vi = 2.00 m/s, when suddenly kinetic friction between M and the surface is triggered (the brakes are switched on). The coefficient of kinetic friction is μk = 0.700. How long does it take for the system to stop if m2 = m1 = 100 kg ? Provide an expression as well as a numerical value.

    2. Relevant equations
    ΣF = ma
    F* = F

    (Newton´s laws)
    3. The attempt at a solution
    I think I did a correctly, but I have some problems solving b.
    I put positive direction downwards to the left.

    a) Firstly I separated the system into 3 systems:

    1) am1 = m1g - T1
    2) am2 = T2 - m2g
    3) This one is the third block, lying on the top of the surface.

    am3 = T1 - T2

    then add together:

    so that: a (m1 + m2 + m3) = m1g - T1 + T2 -m2g + T1 - T2

    a = (m1 + m2)*g / (m1 + m2 + m3)

    b)
    I know that μs = 1.00.
    Fs = nμs = mgμs = 100kg * 9.80 * 1.00 = 980N

    That means that m1g - m2g = 980N
    Then divide by g:

    m1 - m2 = 100kg

    Is that the answer? That the difference in mass has to be less or equal to 100kg?
     
    Last edited: Sep 23, 2016
  2. jcsd
  3. Sep 23, 2016 #2
    b)
    I know that μs = 1.00.
    Fs = nμs = mgμs = 100kg * 9.80 * 1.00 = 980N

    That means that m1g - m2g = 980N
    Then divide by g:

    m1 - m2 = 100kg

    Is that the answer? That the difference in mass has to be less or equal to 100kg?
     
  4. Sep 23, 2016 #3
    I tried c as well:

    vi = 2.00 m/s
    μk = 0.700
    How long does it take for the system to stop?

    Fk = ma
    μkmg = ma

    a = μkg

    v = vi + μkgt

    t = vikg

    t = 2.00/0.700*9.80 = 0.292s

    Correct? Or do I have to use the masses as well? m1 = m2 = 100kg?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Problem: Braking an elevator
  1. Elevator problem (Replies: 3)

  2. Elevator problem (Replies: 7)

  3. Elevator Problem (Replies: 5)

  4. Elevator Problem (Replies: 9)

Loading...