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Homework Help: Question Pertaining To Newton's Laws

  1. Sep 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Figure 1 shows a box of mass m2=1.0 kg on a frictionless plane inclined at angle θ=30°. It is
    connected by a cord of negligible mass to a box of mass m1=3.0kg on a horizontal frictionless
    surface. The pulley is frictionless and massless.
    (a) Choose the direction of your coordinate system.
    (b) Identify all forces acting on m1and m2.
    (c) Write down Newton’s 2nd law for m1 and m2.
    (d) If the magnitude of horizontal force F is 2.3 N, what is the tension T in the connecting cord?
    (e) What is the largest value the magnitude of F may have without the cord becoming slack?

    2. Relevant equations

    3. The attempt at a solution

    For (a), does question seem to ask me to draw the coordinate system I want to employ, and label the axes?

    For (e), I am sort of confused as to how to answer the question. From the diagram, the direction suggests that the question i impossible to answer, for it acts in the direction of the tension force. Wouldn't F have to be zero, because, if it weren't, then it would cause m1 to accelerate even more, under the influence of both T and F, and m1 would push into the string, causing it to bunch up, whereby it won't be taut any longer.

    Attached Files:

    • 001.jpg
      File size:
      17.8 KB
    • 002.jpg
      File size:
      14.9 KB
    Last edited: Sep 1, 2013
  2. jcsd
  3. Sep 1, 2013 #2
    a)Correct- so what do you choose?
    A diagram would be nice...
  4. Sep 1, 2013 #3
    I am terribly sorry--I forgot to upload the documents. Hold on, in one moment they shall be uploaded.
  5. Sep 1, 2013 #4
    The vector for T is incorrect.
  6. Sep 1, 2013 #5
    Which tension force is incorrect?
  7. Sep 1, 2013 #6

    Doc Al

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    Staff: Mentor

    They also want you to choose a direction for your coordinate system, so that you have consistent sign convention in your equations.

    Hint: What is the tension in the string when it goes slack?
  8. Sep 1, 2013 #7
    Are my sign conventions not consistent? The coordinate system for m1 is positive going to the right and going upwards; and the coordinate system for m2 is positive when going to the right and upwards.

    Hmm, I suppose it would be T = 0, but I can't justify why.
    Last edited: Sep 1, 2013
  9. Sep 1, 2013 #8

    Doc Al

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    Staff: Mentor

    Sounds good to me. (Your diagram is so faint that it is difficult to read.)

    For a rope to be taut means that there is tension in it. No tension, it goes slack.
  10. Sep 2, 2013 #9
    So, the string becomes slack when the condition of F = -T is met?

    Also, Enigman had said that my tension force was incorrect, which one, is what I ask?
  11. Sep 2, 2013 #10

    Doc Al

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    Staff: Mentor

    No, the string being slack means that T = 0.

    Not sure what he meant. Your first attachment is so faint, that I cannot read your equations. Just type them out.
  12. Sep 2, 2013 #11
    Then how am I to answer question e); of what consequence does the force F have on the tension force? In the second picture the force F is directed to the right, wouldn't any force directed to the right cause the string to go slack? Following this line of reasoning, wouldn't F have to be zero?

    I don't have any equations written on either page I uploaded; I think he was referring to the direction which I have assigned to the tension force is wrong.
  13. Sep 2, 2013 #12
    When does the T become zero, what direction does F need to be in?
    Nevermind, got confused between the two FBDs.:redface:
  14. Sep 2, 2013 #13
    It needs to be opposite to the direction that it is written. Does the string become slack when F < -T?
  15. Sep 2, 2013 #14
    Think of it this way- A horse(m2) is pulling a cart (m1)- In what case would the string connecting them will slack given that a car is pushing it?
  16. Sep 2, 2013 #15
    If the horse began moving backwards, towards the cart, the string would become slack.
  17. Sep 2, 2013 #16
    Yes, but lets assume (m2) Horse doesn't change the direction of motion. Think how much force the car would exert on the cart so that string is slackened- think intuitively.
    BTW I'm still not sure if the third FBD is completely correct...gravity components seem to be off...can't make it out properly though...too faint.
  18. Sep 2, 2013 #17
    Newton's second law for m2:

    [itex]\sum F_x = mg \cos (30^{\circ}) - T = m_2a_2[/itex]

    [itex]\sum F_y = N - mg \sin(30^{\circ})[/itex]

    Do those appear correct?


    I am confused, as you are referring to a car and also a cart.
  19. Sep 2, 2013 #18
    Oh, everything is correct except the gravity components. Shouldn't it be cos(600) or sin(300) in Fx? Check m2's first fbd and the second one.
    Also won't the tension disappear if the m2 starts accelerating more than the m1?
  20. Sep 2, 2013 #19
    1>If string isn't taut T doesn't act.
    2>How much acceleration(/force F) is needed to slacken the string?
    -What about F? where is it?
  21. Sep 2, 2013 #20
    Yes, the tension would disappear. The direction of the acceleration of m1 would be towards the right, correct?
  22. Sep 2, 2013 #21
    As of now, I am feeling rather uncertain about this entire problem.
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