A balanced lever has two weights on it, the first with mass $2 k g$ $2 kg$ and the second with mass $3 k g$ $3 kg$ . If the first weight is $7 m$ $7 m$ from the fulcrum, how far is the second weight from the fulcrum?

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Answer 1 · 2016-09-20T07:42:03

$x = \frac{14}{3} m$ $x=14/3" "m$

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$G_{1} = m_{1} \cdot g = 2 g$ $G_1=m_1*g=2g$

$G_{2} = m_{2} \cdot g = 3 g$ $G_2=m_2*g=3g$

$G_{1} \cdot 7 = G_{2} \cdot x$ $G_1*7=G_2*x$

$2cancel(g)*7=3cancel(g)*x$

$14=3x$

$x=14/3" "m$

A balanced lever has two weights on it, the first with mass 2 kg 2kg2 kg and the second with mass 3 kg3kg3 kg. If the first weight is 7 m7m 7 m from the fulcrum, how far is the second weight from the fulcrum?