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Tension (Angles and Inclines)

  1. Feb 18, 2015 #1
    1. The problem statement, all variables and given/known data
    A 15kg block sitting on a 22(degree) incline is held stationary by a string as shown. The coefficient of friction between the block and the surface of the incline is 0.12.

    What is the minimum tension in the string?

    2. Relevant equations
    F=mg

    3. The attempt at a solution
    Drew a free-body diagram.


    http://postimg.org/image/dxczxynad/
     
  2. jcsd
  3. Feb 18, 2015 #2
    Not certain, but if I calculate the Fgparallel portion, will I be able to calculate my Tension Force?
     
  4. Feb 18, 2015 #3

    Nathanael

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    Yes that will be necessary, but then what? Consider a free body diagram, what are all the forces acting on the block? Two of them are variable.

    Edit:
    I didn't see you already drew one. Which of the forces are variable? (Hint: their sum is not variable)
     
  5. Feb 18, 2015 #4
    Well, I notice that both Tension Force and Force Friction are along the same path, therefore I believe those are the two variables?
     
  6. Feb 18, 2015 #5
    It is stationary thus the net sum of forces acting it is zero.
     
  7. Feb 18, 2015 #6

    Nathanael

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    Well, yes. But not for your reasoning. The reason is that the force of static friction is variable. μsFn only represents the maximum force that static friction can apply, but it is possible for static friction to apply less or even no force at all.

    You're asked to find the minimum tension in the rope. When do you think the force of tension will be minimum?
    When the static friction is zero? When the static friction is maximum? Or somewhere in between?
     
  8. Feb 18, 2015 #7
  9. Feb 18, 2015 #8


    I noticed you mentioned static friction.

    Since the object is at rest, I believe that the static friction is somewhere in between.
     
  10. Feb 18, 2015 #9

    Nathanael

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    It can be somewhere in between. Or it can be maximum. Or it can be zero. All of these are valid possibilities. But for each of these options, the force of tension will be different.
    When will the force of tension be minimum? Are you saying tension is minimum when the force of friction is somewhere in between it's minimum (zero) and it's maximum?
     
  11. Feb 18, 2015 #10
    Tension force is minimum when force of friction is at it's maximum.
     
  12. Feb 18, 2015 #11

    Nathanael

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    Good. So now find the minimum force of tension. (That is, find the tension needed to keep the object stationary when static friction is applying it's maximum force.)
     
  13. Feb 18, 2015 #12
    Okay, I think I may be getting what you mean, however I end up with a diagram like this. Is there anything in particular that I am doing incorrect?

    http://postimg.org/image/nujh0ea99/

    http://postimg.org/image/y9ir71kwn/
     
  14. Feb 18, 2015 #13

    Nathanael

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    How do you get Fgravity parallel=64.7 N and Fgravity perp.=147 N?
     
  15. Feb 19, 2015 #14
  16. Feb 19, 2015 #15

    Nathanael

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    Ok you have the right idea, but 64.7 is not the right number.

    Can you finish solving for the minimum tension now?
     
  17. Feb 19, 2015 #16
    147cos68 = 64.70102431 N
    ^ This is my adjacent, a.k.a. my Gravitational Force parallel. Should the opposite of it not equal to both the Tension Force and Frictional Force combined? I thought that is how equilibrium works.

    I am having difficulty finding the correct formula to solve. My assumptions are that (0.12 Mew * Normal Force) is my maximum Frictional Force? The issue is also found when I try to find the adjacent for the 22 degree triangle.
     
  18. Feb 19, 2015 #17

    Nathanael

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    This is not true... You need to make sure your calculator is in "degrees mode" and not radians.

    Yes this is correct.

    Yes this is also correct.

    I'm not sure I understand what you're confused about. Are you having trouble finding the normal force?
     
  19. Feb 19, 2015 #18

    Oh my heavens...

    Mr. Nathanael, I am terribly sorry... This entire time I was using radians!
    Brief overview.

    Force of Gravity Parallel = 147cos68 = 55.0672 N
    Force of Gravity Perpendicular = 147cos22 = 136.2960 N
    Force of Friction = (Mew * Normal Force) = (0.12)(136.2960) = 16.35552

    MINIMUM TENSION FORCE = Force of Gravity Parallel - Force of Friction = (55.0672) - (16.35552) = 38.71168 = 39 N
     
  20. Feb 19, 2015 #19

    Nathanael

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    Good job, it is correct.
     
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