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anonymous
 4 years ago
Q.
anonymous
 4 years ago
Q.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1350257258107:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Two 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
 4 years ago
Best ResponseYou've already chosen the best response.0Let's do c) first dw:1350257504270:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1350257622447:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \theta = \arctan \left( F \over mg\right) \\ \text{If F} \to \infty , \text{, you know where will be equilibrium position}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0 end of part 'c' and 'd' 

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for part 'a' and 'b', this is pretty weird dw:1350257959167:dwdw:1350258004726:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw: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
 4 years ago
Best ResponseYou've already chosen the best response.0When F = 0, equilibrium angle = 0 When F=Inf, equilibrium angle = 90 dw:1350258454034:dwdw:1350258475935:dw
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