A balanced lever has two weights on it, the first with mass $8 k g$ $8 kg$ and the second with mass $24 k g$ $24 kg$ . If the first weight is $6 m$ $6 m$ from the fulcrum, how far is the second weight from the fulcrum?

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Answer 1 · 2016-07-09T07:30:36

$x = 2 m$ $x=2" "m$

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$F_{1} \cdot ¯ ¯¯¯¯ ¯ A B = F_{2} \cdot ¯ ¯¯¯¯ ¯ B C$ $F_1* bar (AB)=F_2* bar (BC)$

$F_{1} \cdot 8 = x \cdot F_{2}$ $F_1*8=x*F_2$

$x = \frac{8 \cdot F_{1}}{F_{2}}$ $x=(8*F_1)/F_2$

$plug F_{1} = 6 \cdot g; F_{2} = 24 \cdot g$ $"plug "F_1=6*g" ; "F_2=24*g$

$x=(8*6*cancel(g))/(24*cancel(g))$

$x=48/24$

$x=2" "m$

A balanced lever has two weights on it, the first with mass 8 kg 8kg8 kg and the second with mass 24 kg24kg24 kg. If the first weight is 6 m6m 6 m from the fulcrum, how far is the second weight from the fulcrum?