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Angle Made By a Pendulum

  1. Jan 29, 2016 #1
    1. The problem statement, all variables and given/known data
    WP_20160129_22_47_05_Pro.jpg

    2. Relevant equations


    3. The attempt at a solution
    Using parallel lines I got the angle as theta.
     
  2. jcsd
  3. Jan 29, 2016 #2

    Suraj M

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    Identify the forces on the pendulum bob through components of mg using theta
     
  4. Jan 29, 2016 #3
    The components of weight are mgcostheta and mgsintheta and there is tension acting in the string.
     
  5. Jan 29, 2016 #4

    Suraj M

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    Could you draw a diagram to represent those forces on the bob,?
     
  6. Jan 29, 2016 #5
    WP_20160129_23_43_31_Pro.jpg
     
  7. Jan 29, 2016 #6

    Suraj M

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    Since it's in free fall along the plane I think you should take a pseudo force along that line of motion(##mg\sin\theta##)
    Extend the length of the pendulum and label ##\alpha## I think you can proceed from there.
     
  8. Jan 29, 2016 #7

    haruspex

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  9. Jan 30, 2016 #8
    The tension in the string has components too. So T sin alpha=mg
    Sin alpha = 1
    So, alpha is 90.
    Thank you!
     
  10. Jan 30, 2016 #9

    haruspex

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    Right answer, but I am baffled by your path to it.
    I assume you are taking alpha as the angle between the string and the roof. If so, T sin alpha is not mg. And to perform your next step, you somehow had to have that T=mg. Where did that come from?
     
  11. Jan 30, 2016 #10
    Alpha is the angle between the string and the roof. The tension in the string will be mg, wouldn't it? Because it'll be the mass of the bob and g will be acting on it.
     
  12. Jan 30, 2016 #11
    Alpha is the angle between the string and the roof. The tension in the string will be mg, wouldn't it? Because it'll be the mass of the bob and g will be acting on it.
     
  13. Jan 30, 2016 #12

    haruspex

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    Two forces act on the bob, the tension in the string and mg. You don't yet know what direction the tension is in. Also, the bob is accelerating, so the net force is not zero.
    What you do know is that there is no acceleration perpendicular to the plane, so the forces must balance in that direction. But that still leaves you with two unknowns, T and alpha. So you need to use the known downplane acceleration of the system.
     
  14. Jan 31, 2016 #13
    But won't the vertical component of tension be T sin alpha anyway? And the vertical components need to be balanced. But then which force would I equate it to?
     
  15. Jan 31, 2016 #14

    haruspex

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    The component of the tension perpendicular to the roof will be T sin alpha, but that is not vertical.
    The vertical components must balance if the acceleration has no vertical component, but it will have.
     
  16. Jan 31, 2016 #15

    Suraj M

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    The component of gravitational force perpendicular to the inclined plane? Did you consider that? Relate that to the tension
     
  17. Feb 4, 2016 #16
    Can I do this:
    mgsin theta + Tcos alpha - mgsin theta =0

    mgsintheta is the horizontal weight component and the horizontal component of T acts in the same direction as it. The -mgsintheta is from the pseudo force that acts on the block.

    So, Tcos alpha = 0
    Cos alpha =0
    Alpha = 90
    ]
     
  18. Feb 4, 2016 #17

    haruspex

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    That works if you change all occurrences of 'horizontal' to .... what?
     
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