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El_Psy_Congroo!!

  • 2 years ago

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  1. El_Psy_Congroo!!
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    |dw:1350257258107:dw|

  2. El_Psy_Congroo!!
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    Two constant forces are acting on that point, the resultant is \[ \sqrt{(mg)^2 + F^2} \] and the angle is \[ \theta = \arctan \left ( F \over mg\right)\]

  3. El_Psy_Congroo!!
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    Let's do c) first |dw:1350257504270:dw|

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    |dw:1350257622447:dw|

  5. El_Psy_Congroo!!
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    \[ \theta = \arctan \left( F \over mg\right) \\ \text{If F} \to \infty , \text{, you know where will be equilibrium position}\]

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    ------------- end of part 'c' and 'd' ----------------

  7. El_Psy_Congroo!!
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    for part 'a' and 'b', this is pretty weird |dw:1350257959167:dw||dw:1350258004726:dw|

  8. El_Psy_Congroo!!
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    |dw:1350258159801:dw| \[ H = L - L \cos(2\theta) = L( 2 \sin^2 \theta ) = 2L \sin^2 \theta = 2L{\tan^2 \theta \over 1 + \tan^2 \theta }\] Just put the values of \( \tan (\theta)\)

  9. El_Psy_Congroo!!
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    When F = 0, equilibrium angle = 0 When F=Inf, equilibrium angle = 90 |dw:1350258454034:dw||dw:1350258475935:dw|

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