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Newtons law

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data

    The problem statement, all variables and given/known data[/b]

    am i right? please help me .. thnx

    The problem statement, all variables and given/known data
    the blocks shown in the figure are connected by flexible, inextensible cords passing over frictionless pulleys. at A, the coefficient fiction is 0.30 while at B, it is 0.40. compute the magnitude and direction of the friction forces and each block.

    2.jpg

    2. Relevant equations
    for block A=
    T-mgsin53.13-Fmax=0
    Fmax=0.3(300cos53.13)

    for block B=
    T-mgsin36.87-Fmax=0
    Fmax=0.40(200cos36.87)


    3. The attempt at a solution
    for A:

    Fmax=μk(m1)gcos53.13
    54=μk(m1)gcos53.13
    μk=0.03

    T-300(9.81)sin53.13-μk(300)9.81cos53.13=(300)a eq(1)
    200g-T=200a eq(2)

    subtitute (1) and (2) to find a

    a=-3.79

    for block B:

    Fmax=μk(m2)gcos36.87
    64=μk(m2)gcos36.87
    μk=0.04



    T-200(9.81)sin36.87-μk(200)9.81cos36.87=(200)a eq(1)
    300g-T=300a eq(2)

    subtitute (1) and (2)

    to find a

    a=0.2


    AM i DOiNG RiGHT ?? please comment .. im lack of knowledge ..
     
  2. jcsd
  3. Sep 26, 2011 #2
    help me please ! im confuse what would be the fomula i`l use
     
  4. Sep 26, 2011 #3

    gneill

    User Avatar

    Staff: Mentor

    A couple of things. First, the weights of the blocks are specified in pounds (lbs), so unless you convert to Newtons (and kilograms for mass) you can't use g = 9.81 m/s2 because the units won't match. Perhaps you might consider switching the units to kg and N before proceeding, or use g = 32.2 ft/s2 and associated other units of the English system in your calculations.

    Second, before writing the formulas that will describe the Free Body Diagrams for each block you should determine which (if any) block will rise and which will fall. This will let you ascertain the directions to assign to the frictional forces. To do this, imagine that you hold the 300lb block in place and banish all friction. The system will be held static by your holding that block in place. What is the tension in the rope attached to the 300lb block? Is it sufficient to move the block upward on its slope when you release it, or will it want to fall?
     
  5. Sep 26, 2011 #4
    thnx .. block A = 136.08 kg
    B=90.72 kg

    i already draw the free body block A will fall and block be will go upward..

    the motion going to left.
    what would be posible formula to use in this problem? are my solution right?
     
    Last edited: Sep 26, 2011
  6. Sep 26, 2011 #5
    i need to know if my formula is right or wrong? f wrong what part? or at begining im wrong ? what wod be the possible formula i used,,, thnx alot
     
  7. Sep 26, 2011 #6

    gneill

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    Staff: Mentor

    I don't see where you've accounted for the fact that the accelerations of the two blocks will be different (thanks to the pulley geometry) and how the tensions in the two ropes will be different (again thanks to the pulley geometry). How are they related?

    With these relationships in hand, if you then write equations for the two accelerations you'll have sufficient information to solve the simultaneous equations.
     
  8. Sep 26, 2011 #7
    i solve it again through dis solution..

    so :

    T-m1gsin53.13-μkm1gcos53.13=m1a
    eq 1

    T-505.44=90.72a

    eq2

    T-m2gsin36.87-μkm2gcos36.87=m2a

    so T=577.27
    a=-3.42


    for 2nd block:

    T-m2gsin36.87-μkm2gcos36.87=m2a

    eq 1 : T-505.50=90.72a

    T-m1gsin53.13=μkm1gcos53.13=m1a

    eq2 : T-1043.93=136.08a

    so T=571.36

    a=0.73


    am i doing right?? please help
     
  9. Sep 26, 2011 #8

    gneill

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    Staff: Mentor

    No, it doesn't look right to me. You're still not relating the separate accelerations and tensions via the pulley constraints. Also, I can't understand why you'd be writing two FBD equations for each block and using the properties of the other block's situation to do so (its slope angle and frictional coefficient).

    Write one FBD equation for each block. It will encompass just the forces acting on that block. For block 1 leave the tension as T1 and the acceleration as a1. Do the same for block 2, but here the tension is T2 and the acceleration a2.

    Then write down the relationships that pulley enforces on a1 and a2, and on T1 and T2. Use these relationships as two more equations. Altogether you'll have four equations: two from the FBD's for the blocks and the two 'pulley relationship' equations. Simple substitution should allow you to solve for the accelerations and tensions.
     
  10. Sep 26, 2011 #9
    sory im weak here in physics ..

    hihi! this?
    μm1g-T1=m1a1
    m2gsin(teta)-T2=m2a2

    so

    a=(g(μm1-m2))/m1+m2

    last try .. hihihi sorry ..
     
  11. Sep 26, 2011 #10

    gneill

    User Avatar

    Staff: Mentor

    Nope.

    Why don't you focus on just one block for now. Take the left hand one, with mass m1 sitting on a slope of [itex]\theta 1 = 36.87°[/itex], friction coefficient [itex]\mu1[/itex], and with rope tension T1. That should be sufficient information to write the one FBD equation.
     
    Last edited: Sep 26, 2011
  12. Sep 26, 2011 #11
    for A:

    T1=136.08sin53.13/sin36.87

    B:

    T2=90.72sin36.87/sin53.13

    maybe il copy later im weak .. grr!!! sorry sir thnx for intertaining
     
  13. Sep 27, 2011 #12
    for block A

    Fmax=0.30(136.08)cos53.13
    Fmax=24.5
    T=300(9,81)sin53.13
    T=2354.4
     
  14. Sep 27, 2011 #13
    For B
    Fmax=0.4(90.72)cos36.78
    Fmax=21.77
    T=90.72sin36.87
    T=1177.20

    is this right?
     
  15. Sep 27, 2011 #14

    gneill

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    Staff: Mentor

    No. I don't understand why you're mixing angles from the opposite blocks in the equations for a given block. For example, for block A, the 136kg block sitting on the 36.87° slope, why do you have cos(53.13) and sin(53.13) in its FBD equation? When you write the equation for one block, consider only the forces and conditions that apply to that one block. For block A leave the acceleration as a1 and the tension as T1. DO NOT TRY TO FIND NUMERICAL RESULTS FOR THEM AT THIS STAGE! You don't have enough information yet. Note that a1 DOES NOT equal a2, and T1 DOES NOT equal T2.

    You cannot solve for the accelerations and tensions until you have resolved the pulley's influence: You must state the relationships between the accelerations and tensions that the pulley creates.
     
  16. Sep 27, 2011 #15
    for A:
    T-m1gsin53.13-μkm1gcos53.13=m1a
    m1g-T=m2a

    (T-1065.5)/136.08=a
    (1334.94-T)=90.72a

    T=1227.4
    a=1.19

    hehe! im trying .. my memory is low :)
     
  17. Sep 27, 2011 #16
    last attempt to try .. hihi
     
    Last edited: Sep 27, 2011
  18. Sep 27, 2011 #17

    gneill

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    Staff: Mentor

    The accelerations of the blocks are not the same. You must use separate variables for each. The tensions in the ropes attached to each block are not the same. You must use separate variables for each.

    Use a1, T1, a2, T2 for these variables. Write the FBD equations for each block (separately) using these variables for the unknowns. Do not try to combine the equations yet. You might want to use symbols only for now (no numerical values). In which case, use u1 and u2 for the friction coefficients and q1 and q1 for the angles.

    Here's a new diagram that may help:

    attachment.php?attachmentid=39321&stc=1&d=1317103032.gif
     

    Attached Files:

  19. Sep 27, 2011 #18
    3-1.jpg

    teta=tan^-1(.3)
    teta=16.70

    sine law:
    T/sin53=136.08/sin36.87
    T=181

    B::
    teta=tan^-1(.4)
    teta=21.80


    is this right?

    T/sine36.87=90.72/sine53.13
    T=68.04
     
    Last edited: Sep 27, 2011
  20. Sep 27, 2011 #19

    PeterO

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    Homework Helper

    Diagrams in last two posts represent the two angles like 45 degrees, so no hints as to which functions and angles you should be using.

    Did you recognise that the original situation is based on a 3-4-5 triangle, so the sine and cosine values are easy to compute.
     
  21. Sep 27, 2011 #20
    Yes, the frictional force in the given problem, as presented, is independent of acceleration or velocity of either block, but only dependent upon the direction of the normal force for each.

    However, *as presented*, the velocity of each block is unknown, and therefore the direction of the direction of the frictional force is ambiguous.
     
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