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Pulley's-have test coming up soon-please help

  1. Mar 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Two objects of masses, m1=10 kg and m2=20kg are connected by a cord over pulley. m1 sits on a horizontal frictionless table. The system is released and m1 accelerates to the right while m2 accelerates downward.


    2. Relevant equations
    a.) Calculate the magnitude of the normal force exerted by the table on m1.

    b.)By applying Newton's 2nd law to m1 and m2, find the two equations needed to determine the acceleration and the tension in the cord.

    c.)using the equations from part a, calculate the acceleration and the tension
     
  2. jcsd
  3. Mar 19, 2009 #2
    You do not seem to have included any work, but nonetheless I will give you a push in the right direction.

    For part a) I assume you know that the normal force presses up on the object with the same force that the object presses on the table (its weight). In Newtons of course~

    For part b) I believe you should first draw free-body diagrams for each, then work on finding the [tex]\sum[/tex]Fx and [tex]\sum[/tex]Fy.

    That should give you an idea of what to do next.

    Let us know how things move along and good luck on your test. :zzz: I'm sleepy..
     
    Last edited: Mar 19, 2009
  4. Mar 20, 2009 #3
    Okay here goes my attempt:

    a.) The object does not move on the vertical axis, so Fn=m*g

    9.8*10=98
    Normal force acting on object one is 98N

    b.)

    For object one- The only force acting on the x-axis is the tension.
    So: T=m*a

    For object two- I came up with

    -m2g+T=m*a

    c.) So, For object one T=m*a

    For object two: T-mg=m*a

    I used the substitution method and got a= -19.9 m/s^2 and T=-196N

    I'm pretty sure I did something wrong, but I can't figure it out!!
     
  5. Mar 24, 2009 #4
    Hmm. From the way you word things it seems like you cannot check this answer online.. I have a bad track record with these types of problems as recently as last week, but I will try to give some input.

    First of all, consider the direction of each acceleration. Are both acceleration in the positive direction?

    The equation you used to solve for your answer was.

    T-m2g=m2a

    substituting in gave you.

    m1a-m2g=m2a

    However these equations do not take into account the direction of each acceleration.

    The acceleration of m2 is negative along the y-axis, see what you can change within the equation to show that the acceleration is negative along the y-axis
     
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