The answer is Hx = ½ Wsin θ cos θ
The explanation for this is:
Analyzing the torques on the bar, with the hinge at the axis of rotation, the formula would be: ∑T = LT – (L/2 sin θ) W = 0
So, T = 1/2 W sin θ. Analyzing the force on the bar, we have: ∑fx = Hx – T cos θ = 0Then put T into the equation, we get:∑T = LT – (L/2 sin θ) W = 0

4 0

3 years ago

Square root of 32 x square root of 1 over 18

lisov135 [29]

<h3>Solution:</h3>

$\sqrt{32} \times \sqrt{ \frac{1}{18} } \\ = \sqrt{32} \times \frac{ \sqrt{1} }{ \sqrt{18} } \\ = \sqrt{32} \times \frac{1}{ \sqrt{18} } \\ = \frac{\sqrt{2 \times 2 \times 2 \times 2 \times 2}}{ \sqrt{2 \times 3 \times 3} } \\ = \frac{ \sqrt{ {2}^{2} \times {2}^{2} \times 2} }{ \sqrt{2 \times {3}^{2} } } \\ = \frac{2 \times 2 \times \sqrt{2} }{3 \times \sqrt{2} } \\ = \frac{4 \times \sqrt{2} }{3 \times \sqrt{2} } \\ = \frac{4}{3}$