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Kinematics and dynamics

  1. Jun 29, 2009 #1
    The weights of the blocks A, B and C, shown in figure worth, 89N, 44,5 N and 44,5 N, respectively. The blocks are initially at rest. Knowing that B travels 2,44 m in 2s, determine (a) the module of F and (b) the tension on the rope AD. Discard the masses of the pulleys and friction.

    (a) 7,37 N
    (b) 92,7 N

    My solution

    Block B
    ZFy = mB . aB
    T - F - mB.g = mB . aB

    Block C
    ZFy = mC . aC
    -mC . g + T = mC . aC

    aB = -aC

    I can not find the answer of the book

    Attached Files:

  2. jcsd
  3. Jun 30, 2009 #2


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    Using the formula s = 1/2*at^2, find the acceleration of mb.
  4. Jun 30, 2009 #3


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    Hi Apprentice123! :wink:
    Nooo … not this time!

    because … ? :smile:
  5. Jul 1, 2009 #4
    Because ? I do not know
    I have:

    x = 1/2 * aB * t^2
    aB = 1.22 m/s^2

    Block B
    T - F - PB = mB * aB
    F = T - 50,034

    Block C
    -PC + T = mC * aC
    T = 44,5 + 4,5361aC

    How to calculate aC?
  6. Jul 1, 2009 #5


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    For block B, it should be
    F + mB*g - T = mB*aB
  7. Jul 2, 2009 #6
    Yes, thanks. How do I calculate aC?
  8. Jul 2, 2009 #7


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    Hi Apprentice123! :smile:
    General tip: if you can't see clearly what to do, then give everything names (or letters) :wink:

    call the lengths of the ropes above A B C and D a b c and d, respectivley.

    Then the height of B is -(b + d), also b + c is constant, and so is a + d.

    Carry on from there … :smile:
  9. Jul 2, 2009 #8


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    What is this module, first time for me? Don't you think it is the resultant force acting on B?
  10. Jul 2, 2009 #9


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    No. If you observe the weights of A, B and C , You can see that they are in equilibrium. To start the motion, a force F is applied on B.
  11. Jul 3, 2009 #10


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    So the accelerations of B and C will differ due to the fact that the acceleration of pulley D "adds to" the motion of B and "subtracts from" the motion C?
  12. Jul 13, 2009 #11
    I have a very similar question that I'm having problems solving,
    For mass B,
    [tex]\downarrow[/tex]F + 44.5 - T = 4.45a
    and a=1.22ms-2

    For C,
    [tex]\uparrow[/tex] T-44.5 = ma'

    At first I thought a=a' and I tried working with it but I get a different answer,and after reading the other posts,I'm not sure they(the accelerations) are the same,but I don't understand why they are not either and how I'm supposed to calculate it.

    Hope my problem's clear and I hope someone can help me
    Thanks in advance
  13. Jul 13, 2009 #12


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    Hi leena19! :smile:

    See post #7 …

    The height of B is -(b + d), and so the accleration of B is -(b'' + d''), while for C it's … ? :wink:
  14. Jul 13, 2009 #13
    Hello tiny-tim

    I've seen it,sadly I don't understand it :(

    I understand ,why b+c and c+d is constant.Also,why a+b+d and a+c+d is constant,cause the length of the string's constant.
    but i don't get why the height of B is negative(b+d),are we taking displacements here?
    i don't know.is it -(c+d) or (c+d) ? and you get aB = -(b'' + d'') ,by double differentiaion,is it?
    When i use,
    a+b+d , i get aA + aB = 0,and aA = -aB
    and for a+c+d,i get aA = -aC ,which can't be right,right?
    I've never dealt with 2 pulley problems before,but I've done alot of single pulley problems,but we've never had to consider the heights there to find the acceleration,so I'm a bit lost here.

  15. Jul 14, 2009 #14
    This is an old thread I found on double pulleys


    I understand here why the acceleration is doubled for m1,as explained in the thread,but
    I don't get why we can't apply the same principle here.
    I also can't understand how the accelerations of B and C are different,in this system of 3 pulleys?and what the heights of the weights' got to do with it?
    I think the acceleration of A is 1/2 that of B &/or 1/2 the acceleration of C?

    I hope someone can explain to me the motion of this pulley system.

    Thank you
    Last edited by a moderator: Apr 24, 2017
  16. Jul 14, 2009 #15


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    Accelerations are vectors, and since the motion here is one dimensional one can add/subtract the various accelerations that the components in this system is experiencing. This also follows from Newton's second law, that is each force in the resultant equation for each mass has its own contribution to the resultant acceleration.

    That being said one can see that D and A will share the same acceleration. D's acceleration is downwards in this situation. B and C would have had the same acceleration if it were not for the acceleration of D.

    B accelerates downwards and C upwards. This mean that their common acceleration will be altered as follows:

    The acceleration of D will subtract from their common acceleration for C and add for the acceleration of B.
  17. Jul 14, 2009 #16
    Hi all,

    I tried solving this. And got a different answer. Below is what I did.

    Let us assume that acceleration of a block is positive in downward direction (as B moves down). Let the tension in the rope A-D be T1 and that in rope BC be T2.

    Note that the system is an accelerating system. This implies that pulley D is going down with acceleration aA. Also with respect to D, accelerations of B and C need to be equal but in opposite directions. If accelerations of B and C wrt to D are aBD and aCD respectively, then

    aB=aA+aBD aB is acceleration of B with respect to fixed point of pulleys between A and D
    aC=aA+aCD aC is acceleration of C with respect to fixed point of pulleys between A and D

    Hence, aC=aA-aBD

    For block B,

    For block C,

    For pulley D (as its massless),

    For block A,
    T1-mA*g = mA*aA

    Now, we have been given the values of the weights in N (i.e. in the form of mX*g). Also given, is the displacement of B and the duration for the displacement.

    aB=2*s*t^2 gives us aB=1.22 m/s^2

    assuming g=10 m/s^2 gives us mA=8.9 kg, mB=mC=4.45 kg, we get for block B,

    From here on, mA=2m and mB=mC=m. Adding equations obtained for block B and block C,
    F + 2*m*g - 2*T2 = 0 as aCD = - aBD
    F + 2*m*g = T1 ------------- (1)

    Subtracting them gives,
    F = 2*m*aBD ------------- (2)

    From equation for block A we have,
    T1 - 2*m*g = 2*m*aA
    T1 = 2*m*aA + 2*m*g -------------- (3)

    Using (3) in (1),
    F = 2*m*aA

    Compare this (2),
    2*m*aA = 2*m*aBD
    aBD = aA (:D moment)

    As aB = aA + aBD,
    aA = 0.5 * aB = 0.61 m/s^2

    F = 2*m*aA = 54.29 N
    T1 = F1 + 2*m*g = 143.29 N

    If someone can find my mistake kindly assist. Already grateful to those who can help.
  18. Jul 14, 2009 #17
    F = 2*m*aA = 54.29 N
    T1 = F1 + 2*m*g = 143.29 N

    I made a mistake here. The values I got were,

    F = 5.429 N
    T1 = 5.429 + 89 = 94.429 N

    Still very different from the answer. :(
  19. Jul 15, 2009 #18
    Thank you rl.bhat.This is all that I needed to know.
    But just to makesure I've understood it fully, if we were to take the distance interms of d when B is moving downwards,would it be like this?
    b-c+d = 0
    b = c-d
    since -d = +a,
    b = c+a, or
    since aB = aC -AD ,and aD =-aA
    aB = aC + aA
  20. Jul 15, 2009 #19


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    you should use the accelerations aB and aC here, that is the forces are such so that the masses obtained these resultant accelerations - Newton's second law.
  21. Jul 15, 2009 #20
    Still not with you. Could you please explain?
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