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!

Force and motion in Newton's laws

  1. Oct 1, 2011 #1
    two containers are connected by a cord (of negligible mass) passing over a frictionless pulley (also of negligible mass). At time t = 0, container 1 has mass 1.30 kg and container 2 has mass 2.8 kg, but container 1 is losing mass (through a leak) at the constant rate of 0.200 kg/s.

    a) At what rate is the acceleration magnitude of the containers changing at t = 0?
    (b) At what rate is the acceleration magnitude of the containers changing at t = 3.00 s?
    (c) When does the acceleration reach its maximum value?


    [PLAIN]http://img5.imageshack.us/img5/50/q55u.jpg [Broken]

    Take upwards as positive

    Second Newton's law
    m1 : -m1 g + T = m1 a (1)
    m2: -m2 g + t = -m2 a (2)

    (1) - (2)--------> g(m2 - m1) = a (m1 + m2)

    a = g(m2 - m1) / (m1 + m2)


    a. t = 0

    m1 = 1.3kg
    m2 = 2.8 kg

    a = 3.58 m/s^2

    b. t = 3s

    m1 = 1.3 - 3 * 0.2 = 0.7 kg
    m2 = 2.8 kg

    a = 5.88 m/s^2

    c. Ok, I don't know how to do this part. I believe that in order to get max. acceleration, m1 must be = 0, so a will be equal to g ??? is that correct ???
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 1, 2011 #2
    It's correct.
     
  4. Oct 1, 2011 #3

    gneill

    User Avatar

    Staff: Mentor

    They're asking about the rate of change of the acceleration, not the acceleration itself.

    If the masses of the containers were constant then the acceleration would be constant also. But m1 is changing with time. This will cause the acceleration to change over time.

    You have found an equation for the acceleration given the current masses of the containers. That's your: a = g(m2 - m1) / (m1 + m2). If m1 is variable rather than constant then a becomes a function of m1, that is, a(m1) = g(m2 - m1) / (m1 + m2). And m1 itself is a function of time. So make the appropriate substitution, or prepare to use the calculus chain rule...

    Time for a little calculus to find the rate of change of a.
     
  5. Oct 2, 2011 #4
    Uhm, I took Calculus I and II 5 years ago, so I don't think I can remember a lot now, can you please give me more hints on how to solve it ?
     
  6. Oct 2, 2011 #5

    gneill

    User Avatar

    Staff: Mentor

    In a problem, when you see a request for a rate of change of some value you can be pretty sure that differentiation of a function describing that value is going to be involved. This problem is a perfect example of that. Since calculus is frequently needed to solve physics problems, it would be a good idea to review it.

    You have worked out an expression for the acceleration that holds for given masses m1 and m2; For any given pair of m1 and m2 you can calculate the constant acceleration that would result. But you're told that in this case m1 is varying -- it is changing with time as its contents leak out. So m1 becomes a changing variable in your acceleration expression. Write an equation for m1(t). Use this function of time in place of m1 in your acceleration equation. Then you'll have a function for acceleration, a(t).

    The question wants the rate of change of acceleration at some particular times. That means it wants to know what da/dt is for those times.
     
  7. Oct 2, 2011 #6
    Hey but still i guess the third part is right.
     
  8. Oct 3, 2011 #7
    So
    a = g(m2 - m1) / (m1 + m2)

    a = g[m2 - (m1 - 0.2t)] / (m1 - 0.2t + m2)

    da / dt = g d[m2 - (m1 - 0.2t)] / (m1 - 0.2t + m2)]/ dt

    with m2, m1 and g are constant ???
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Force and motion in Newton's laws
Loading...