# Question #1411e

##### 1 Answer
Apr 14, 2016

So the answer is $d \to \frac{3}{2}$

#### Explanation:

Let clockwise be positive
Let anticlockwise be negative

Taking moments about ${F}_{1}$ (Sum of moment must = 0)

$0 = - \left(\frac{W}{8} \times \frac{L}{16}\right) - \left({F}_{2} \times \frac{5 L}{8}\right) + \left(\frac{5 W}{8} \times \frac{5 L}{16}\right) + \left(\frac{W}{4} \times \frac{6 L}{8}\right)$

$\frac{5 L {F}_{2}}{8} = \frac{25 W L}{128} + \frac{3 W L}{16} - \frac{W L}{128}$

$\frac{5 L {F}_{2}}{8} = \frac{24 W L}{128} + \frac{3 W L}{16}$

$\frac{5 L {F}_{2}}{8} = \frac{3 W L}{8}$

$\implies 5 L {F}_{2} = 3 W L$

$\textcolor{b l u e}{\implies {F}_{2} = \frac{3}{5} W}$

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$\textcolor{b r o w n}{\text{Total downward force = total up}}$

$\textcolor{b l u e}{\implies {F}_{1} = W - \frac{3}{5} W = \frac{2}{5} W}$

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Thus the ratio of ${F}_{1} / {F}_{2} = \frac{2 W}{5} \div \frac{3 W}{5} = \frac{2}{3}$

So the answer is $d \to \frac{2}{3}$