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Homework Help: Newton's Laws of motion!

  1. Jan 23, 2009 #1
    1. The problem statement, all variables and given/known data

    A smooth ring A of mass m can slide on a fixed horizontal rod. A string tied to the ring passes over a fixed pulley B and carries a block C of Mass M (=2m) as shown in the given figure. At a instant at the string between the ring and the pulley makes an angle [tex]\theta[/tex] with the rod, (a)Show that, if the ring slides with a speed v, the block descends with speed v cos[tex]\theta[/tex]. (b) With what acceleration will the ring start moving if the system is released from rest with [tex]\theta=~30^\circ[/tex]?

    2. Relevant equations

    Newtons Equations, free body diagram

    3. The attempt at a solution

    Well, this question happens to be from a textbook. And its a solved one too.. The solution which they have given is quite complicated! I thought I'd get a brief explanation of "why and how" over it, and yes an alternative method would be highly appreciated!

    The solution, is like this (as given in the book),

    Suppose in a small time interval [tex]\delta t[/tex] the ring is displaced from A to A' and the block from C to C'. Drop a perpendicular A'P from A' to AB. For small displacements A'B~ PB (I didnt get this!), since the lenght is constant (?????), we have,
    AB+BC= A'B+BC'
    AP+PB+BC=A'B+BC'
    AP=BC'-BC=CC' (as A'B=PB)

    AA'[tex]cos\theta[/tex]= CC'

    or [tex]\frac{AA'cos\theta}{\delta t}=~\frac{CC'}{\delta t}[/tex]

    Therefor, (velocity of the ring)[tex]cos\theta[/tex]= (velocity of the block)


    Please help :cry:
     

    Attached Files:

    Last edited: Jan 23, 2009
  2. jcsd
  3. Jan 23, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    When the ring moves with a velocity v horizontally, its component along the string is v*cos(theta). Since the length of string between the ring and block is costant, the velocity of the block is equal to v*cos(theta).
     
  4. Jan 23, 2009 #3
    Well, did you have a look at the diagram? It hasnt been approved yet..
     
  5. Jan 23, 2009 #4

    rl.bhat

    User Avatar
    Homework Helper


    Suppose in a small time interval LaTeX Code: \\delta t the ring is displaced from A to A' and the block from C to C'. Drop a perpendicular A'P from A' to AB. For small displacements A'B~ PB (I didnt get this!), since the lenght is constant (?????), we have,

    When A and A(dash) are very close, AB and A(dash)B are nearly equal. If you take BA(dash) equal to BP, AP becomes A*A(dash)cos(theta) and A*A(dash)/t = velocity. The length of the string is ABC = A(dash)BC(dash)
     
  6. Jan 24, 2009 #5
    okay, thats fine with me, I've several more problems over the free body diagrams! I'll post it in mean time! Thank You!
     
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