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Tension in a string between two blocks

  1. Feb 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Two masses M1 = 6.90 kg and M2 = 3.10 kg are on a frictionless surface, attached by a thin string. A force of 48.1 N pulls on M2 at an angle of 31.3° from the horizontal as shown in the figure. Calculate the tension T in the string.

    2. Relevant equations
    T=ma
    a=g(m2/(m1+m2))

    pullbox.gif

    3. The attempt at a solution
    a=g(m1/(m1+m2))
    a=9.81m/s^2(3.10 kg /(6.90 kg + 3.10 kg))
    a=9.81m/s^2(3.10 kg/10 kg)
    a=9.81m/s^2(.31)
    a=3.0411 m/s^2
    T=ma
    T=(6.90 kg)(3.0411 m/s^2)
    T=20.98359 N
     
  2. jcsd
  3. Feb 3, 2015 #2
    So consider the force applied on the total mass. What should the masses accelerate? From there, isolate the system and see the tension needed to accelerate that isolated system by that amount
     
  4. Feb 3, 2015 #3
    The forces that are applied on the mass are F (48.1 N), gravitational force, and the normal force. Did I find the acceleration incorrectly in my work above?
     
  5. Feb 3, 2015 #4

    BvU

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    Hello Litz, welcome to PF :)

    Your first relevant equation is OK, provided you mean M1 for m.

    But your second one comes from elsewhere and doesn't fit here . In this exercise there is no friction with the surface, so the acceleration from gravity doesn't play a role.
    Instead, consider a second equation just like the first one, but for M2. There are two forces working on M2 that are relevant for this exercise. Can you distinguish which two I mean ?
     
  6. Feb 3, 2015 #5
    Is the static frictional force and F (force from the pull) what you are referring to? I have been struggling with determining forces and questions like the one I posted. My TA and professor have not been helpful when explaining the concepts to me.
     
  7. Feb 3, 2015 #6
    You do not have to consider the weight since its frictionless. You just need to consider forces in the the direction of motion

    What do you think is the horizontal acceleration of the whole system? Consider this before breaking it down to look at tension
     
  8. Feb 3, 2015 #7

    BvU

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    Force from the pull is certainly relevant. The other one I meant has to do with T. There's a Newton law action = -reaction: if the connecting wire pulls to the right on M1 with a force T, then it pulls with an equal but opposite force -T to the left on M2.

    The static frictional force is zero on a frictionless surface. The two other forces that work on each of the two blocks are gravity (M1g and M2g, respectively) working downwards, and the normal forces the surface exerts upwards. These simply cancel (the blocks don't accelerate in the vertical direction, so in that diretiction the sum of forces is zero (*) ) .

    So for block 1 we have something with a T and an a. These also appear when we write the equation for block 2. And there F has to show up too, but not the full 48.1 N. Do you know whereI'm trying to go ?
     
  9. Feb 3, 2015 #8
    Not really. I think I might have an idea, but I do not think that it is correct. Since the full 48.1 does not show up, something needs to be subtracted. Am I supposed to find the T=m1a, then subtract that from the equation for block 2?
     
  10. Feb 3, 2015 #9
    I am not sure what is the horizontal acceleration of the whole system. Do I use F=ma to find the acceleration of the whole system?
     
  11. Feb 3, 2015 #10
    The net force on the system will accelerate the mass of the system by that amount proportionally. Other force pairs are internal to the system and do not change the net force acting on our system, only how the forces act. So, yes
     
  12. Feb 3, 2015 #11

    haruspex

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    Yes, as long as you use the correct value for F in that equation.
     
  13. Feb 3, 2015 #12
    I was told earlier that I cannot use 48.1 N as F, so how do I figure out what F is for this problem?
     
  14. Feb 3, 2015 #13
    Read my post earlier about considering force in the direction of motion
     
  15. Feb 3, 2015 #14

    haruspex

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    You wrote:
    So clearly we are discussing horizontal acceleration. What direction of force produces horizontal acceleration?
     
  16. Feb 3, 2015 #15
    The right?
     
  17. Feb 3, 2015 #16
    You're given a force vector and an angle. Break it down :)
     
  18. Feb 3, 2015 #17
    The force would have to be to the right because if it was to the left the blocks would not accelerate...?
     
  19. Feb 3, 2015 #18
    Do I use Fcos(theta) (48.1*cos(31.3))?
     
  20. Feb 3, 2015 #19
    Mhmm. From there, what is the net acceleration? After this, try to think about the forces that are acting on both blocks, the magnitude of the forces that should act on both blocks. If you think about this, you can make your way toward the tension of the rope.
     
  21. Feb 3, 2015 #20
    Is the acceleration 48.1*cos(31.3)=41.0995?
     
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