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Newton Laws Question

  1. Oct 9, 2005 #1
    A block of mass, m1=3.6kg on a frictionless inclined plane of angle 30 degrees is connected by a cord over a massless,frictionless pulley to a second block of mass,m2=2.1kg hanging vertically..
    I cant find the formula or anything.
    any help?
    as i need to find the magnitudes of acceleration for each block (know that they will be the same) and the tension of the cord...
     
    Last edited: Oct 9, 2005
  2. jcsd
  3. Oct 9, 2005 #2
    Try to figure out the forces acting on each of these two blocks. Then find the components of these forces onto the cord direction and apply F=m*a for your system.
     
  4. Oct 9, 2005 #3
    ive tried that. I figure Fn=mass1 * g * cos(30) =35.28...
    should the tension be just T1= m*g*sin(30)? but it doesnt come out!
     
    Last edited: Oct 9, 2005
  5. Oct 9, 2005 #4
    You have [tex]m_2 g[/tex] from the vertically hanging block and [tex]m_1 g sin(30)[/tex] from another one. So the difference between these two components must equal [tex](m_1+m_2)a[/tex].


    BTW. if Fn is the normal component (to the inclined plane) then it must be given by a cos (and not sin as in your post)!
     
  6. Oct 9, 2005 #5
    Yeah but we have two unknowns... a and Fnet... would Fnet=Fn?
    I found fn with =mgcostheata and got 30.6
    found fg for mass 1=mg=35.28, and got Fg(x)=17.65, and Fg(y)=30.6
    isnt T1=m1*g*costheata?
     
    Last edited: Oct 9, 2005
  7. Oct 9, 2005 #6
    what do you mean by Fnet? =Fn=Gn? You have N that must be accounted for!

    The equilibrium conditions are written SEPARATELY onto the two directions (parallel and perpendicular to the cord direction).

    perpendicular) for m2, Gn=N (Fn=Gn-N=0 so an=0) and for m1 you have neither Fn nor N.

    parallel) For the system, Ft is the difference between the forces I given you in my previous post and it must equal (m1+m2)a.

    So the only unknown you have is a.
     
    Last edited: Oct 9, 2005
  8. Oct 9, 2005 #7
    what does m1 g sin 30 =? is it Fg in the x component?
     
  9. Oct 9, 2005 #8
    YES (if your Ox is parallel to the cord direction)
     
  10. Oct 9, 2005 #9
    ok
    i got some of this
    So to find Acceleration= (Mass2 * g - Mass1*g*sin30)/Masstotal

    i know Mass1*g*sin30 is from Tension of mass 1 but where is that other part from
     
  11. Oct 9, 2005 #10

    lightgrav

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    Homework Helper

    NO, you're ignoring the prepositions!

    m1 g sin(theta) is the parallel component of gravity's Force
    (parallel to the string). It is in the opposite direction from the Force by the string Tension (which is pulling the block _up_ the ramp).
    The SUM of these two Forces, applied to the object in your FBD, is what gives m1 its acceleration.
    m2 can be treated as a separate equation ... which will also have T in it.

    Once you put the two equations together, you should end up with the "unbalanced Force" causing (m1 + m2) a .
     
  12. Oct 9, 2005 #11
    BUT WHAT IS TENSION!
    Tension 1 for Mass 1 is T-FgSin(theata)=ma
    Tension 2 for mass 2 is mg-ma
    All i am asking is When your have the "unblance force" the egn it comes out to is (mass of the 2nd object * g - mass of 1st object * g* sin (theata) )/Mass total = acceleation. I know this, thats the answer, it check but I cant account for where mass of the 2nd object * g is COMING from. I know where mass of 1st object * g* sin (theata) and why its negative is coming from (Fg in the xcomponent of mass 1) but not for that part.
    Where does M2 * g come from, what force is that?
    what am i missing, i see the two equations but what is that force that has that equation?
     
  13. Oct 9, 2005 #12
    The second principle must be applied for EACH block separately. Because perpendicularly to the string direction you have (obviously) equilibrium, we'll write the equations along the string direction only. So, if we suppose that m2 is going down,

    for m1: [tex]T-m_1 g sin (30)=m_1 a[/tex]

    and

    for m2: [tex]m_2 g -T=m_2 a[/tex]

    (the same acceleration and equal tensions but opposite)

    You have to add now these 2 eqs. and find a.
     
    Last edited: Oct 9, 2005
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