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Acceleration of a Mass in a Pulley System

  1. Nov 15, 2016 #1
    • Moved from technical forums, so no template
    The block 1 of m1 sheets of mass on an inclined plane without friction at an angle θ to the horizontal. One end of a non-extendable rope with no mass is attached to lock 1. The chain is wound around a non-frictionally mobile pulley P of negligible mass and is also connected to the wall at the base of the inclined plane. Pulley P is connected for a second no-mass non-stretchable cord to block 2 of m2 of mass through a second, frictionless pulley that is secured in place. Block 2 hovers over the end of the inclined plane (see figure). The gravitational acceleration downwards is g.

    What is the acceleration modulus of block 2 after the system is released from rest? Suppose that m2 is large enough that it is moving down. Express your answer in terms of theta for θ, m1, m2, and g.

    A2 =?

    Attempted to perform with equation


    4 * ((4 m_2- m_1 * * (cos (theta) + sin (theta))) / (4 * + m_1 m_2)) * G

    But it shows an error trying to rearrange the formula, but I can not do it right

    Can you help please?
     
  2. jcsd
  3. Nov 15, 2016 #2
    I think we may need a picture, or at the very least a description where blocks don't become locks and ropes become chains and pulleys are apparently mobile but nevertheless attached to walls.
     
  4. Nov 15, 2016 #3
    fafafaf.png
     
  5. Nov 15, 2016 #4
    Ah. Yes, that is much more clear.

    Note that the motion of box 1 must be twice the motion of box 2. That includes displacement, speed, and acceleration.

    The sum of the forces on box 2 give the acceleration of box 2

    m2 g - T2 = m2 a2

    Where T2 is the tension in the top rope. The sum of the forces on box 1 give the acceleration of box 1

    T1-m1 g sin(Q) = m1 a1

    As I noted above

    a1 = 2 a2

    Also from the pulley

    T2 = 2 T1

    Substituting for T1 and a1 the second equation becomes

    T2 / 2 - m1 g sin (Q) = m1 2 a2

    Now use the first equation to eliminate T2

    (m2 g - m2 a2) / 2 -m1 g sin(Q) = m1 2 a2

    Solve for a2 and simplify. QED
     
  6. Nov 16, 2016 #5

    haruspex

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    I know you just want to be helpful, but this is too much handholding for a homework forum. The principle is to offer hints, ask leading questions, point out mistakes, etc.
     
  7. Nov 16, 2016 #6
    Oops. I've been reading "no homework" so often I didn't realize there was a homework forum. I kind of thought this was probably homework, but after some back and forth told myself who am to judge? If he says it isn't homework then I'll believe it. Now you tell me there is a homework forum.

    I am very sorry. I feel like an idiot. Not only not good but probably genuinely not what the OP would have wanted. In my defense, this is the first thing I've seen that looked like homework and I'm not sure how I got to it.

    So, two questions. I went to the same place I always go (just looking at General physics forum questions) where most of the time I see "no homework" warnings. Did I accidentally do something differently, or are questions from different forums linked together at that level, sort of a meta forum? More importantly how do I tell whether I'm on a no homework or homework forum? I mean I've seen the warning, but I hadn't paid attention as to whether it is there all the time. (I thought I was always on just one forum). Is that the best indicator? I see in the thread list that different threads have colors and labels I don't understand. I bet one of those was trying to tell me something. I now see at the top of this thread that I wound up in "Introductory physics". It doesn't show the parent directories (list too long, ellipsis), but browsing back I see I'm deep in a homework sub level. How the heck did I get here?

    Well dang. Sorry again to all.
     
  8. Nov 16, 2016 #7

    haruspex

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    If you look at the top of this thread you will see it has been moved by a moderator, but it does not say when. Maybe it was still in the technical forum when you posted your first, or even second, reply.
     
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