anonymous
  • anonymous
Q.
OCW Scholar - Single Variable Calculus
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1350257258107:dw|
anonymous
  • anonymous
Two constant forces are acting on that point, the resultant is \[ \sqrt{(mg)^2 + F^2} \] and the angle is \[ \theta = \arctan \left ( F \over mg\right)\]
anonymous
  • anonymous
Let's do c) first |dw:1350257504270:dw|

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anonymous
  • anonymous
|dw:1350257622447:dw|
anonymous
  • anonymous
\[ \theta = \arctan \left( F \over mg\right) \\ \text{If F} \to \infty , \text{, you know where will be equilibrium position}\]
anonymous
  • anonymous
------------- end of part 'c' and 'd' ----------------
anonymous
  • anonymous
for part 'a' and 'b', this is pretty weird |dw:1350257959167:dw||dw:1350258004726:dw|
anonymous
  • anonymous
|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 \( \tan (\theta)\)
anonymous
  • anonymous
When F = 0, equilibrium angle = 0 When F=Inf, equilibrium angle = 90 |dw:1350258454034:dw||dw:1350258475935:dw|

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