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Force on a pully system

  1. Sep 30, 2007 #1
    A block of mass m1 = 4.00 kg on a frictionless inclined plane of angle 30.0° is connected by a cord over a massless, frictionless pulley to a second block of mass m2 = 2.40 kg hanging vertically (Fig. 5-41).

    [​IMG]
    Figure 5-41

    (a) What is the magnitude of the acceleration of each block?
    F=ma; F1=M1g F1 = 23.5N F2=M2Gsin(30)=19.6N Ax1=3.9N/4kg=.975m/s^2 thats wrong, the same for the other mass which I just subbed the 4 with the 2.4kg

    (b)What is the tension in the cord?

    the two forces combined 19.6+23.5=43.12N
    also wrong
     
  2. jcsd
  3. Sep 30, 2007 #2

    Dick

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    There are two forces acting on m1, a component of m1*g acting along the incline and the tension T. There are also two forces acting on m2, m2*g and T. Both have the same acceleration a, so write F=ma for each and solve for a and T.
     
  4. Sep 30, 2007 #3
    I understand that, but that still yeilds a force of 3.9N right, in negative X I have M1gsin30 and in positive I have M2g so, Fx1= M2g - M1gsin30 =3.9N and divided by mass, thats still .975m/s^2 right?
     
  5. Sep 30, 2007 #4

    Dick

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    You are just adding together the gravitational forces. I don't see any tension force in there. Write down the sum of the forces on each object and then equate them to m*a. Call the tension T. It's pulling up on both objects. There are two unknowns in the problem, the tension and the acceleration. You need an F=ma equation for both objects to get two equations to solve for them.
     
  6. Sep 30, 2007 #5
    okay I got you, I need to put T in there from the start, I just supposed T was the difference in the forces.
     
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