## El_Psy_Congroo!! 2 years ago Q.

1. El_Psy_Congroo!!

|dw:1350257258107:dw|

2. El_Psy_Congroo!!

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!!

Let's do c) first |dw:1350257504270:dw|

4. El_Psy_Congroo!!

|dw:1350257622447:dw|

5. El_Psy_Congroo!!

$\theta = \arctan \left( F \over mg\right) \\ \text{If F} \to \infty , \text{, you know where will be equilibrium position}$

6. El_Psy_Congroo!!

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

7. El_Psy_Congroo!!

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

8. El_Psy_Congroo!!

|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!!

When F = 0, equilibrium angle = 0 When F=Inf, equilibrium angle = 90 |dw:1350258454034:dw||dw:1350258475935:dw|