Question #1411e

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Question #1411e

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Answer 1 · 2016-04-14T20:32:03

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

Explanation:

Tony B

Let clockwise be positive
Let anticlockwise be negative

Taking moments about $F_{1}$ $F_1$ (Sum of moment must = 0)

$0 = - (\frac{W}{8} \times \frac{L}{16}) - (F_{2} \times \frac{5 L}{8}) + (\frac{5 W}{8} \times \frac{5 L}{16}) + (\frac{W}{4} \times \frac{6 L}{8})$ $0=-(W/8xxL/16) -(F_2xx(5L)/8)+((5W)/8xx(5L)/16)+(W/4xx(6L)/8)$

$\frac{5 L F_{2}}{8} = \frac{25 W L}{128} + \frac{3 W L}{16} - \frac{W L}{128}$ $(5LF_2)/8= (25WL)/128+(3WL)/16-(WL)/128$

$\frac{5 L F_{2}}{8} = \frac{24 W L}{128} + \frac{3 W L}{16}$ $(5LF_2)/8= (24WL)/128+(3WL)/16$

$\frac{5 L F_{2}}{8} = \frac{3 W L}{8}$ $(5LF_2)/8= (3WL)/8$

$\Rightarrow 5 L F_{2} = 3 W L$ $=>5LF_2=3WL$

$\Rightarrow F_{2} = \frac{3}{5} W$ $color(blue)(=>F_2=3/5W)$

'~~~~~~~~~~~~~~~~~~~~~~~~

$Total downward force = total up$ $color(brown)("Total downward force = total up")$

$\Rightarrow F_{1} = W - \frac{3}{5} W = \frac{2}{5} W$ $color(blue)(=>F_1=W-3/5 W = 2/5 W)$

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Thus the ratio of $\frac{F_{1}}{F_{2}} = \frac{2 W}{5} \div \frac{3 W}{5} = \frac{2}{3}$ $F_1/F_2 = (2W)/5 -:(3W)/5 = 2/3$

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

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Question #1411e

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