A balanced lever has two weights on it, the first with mass $5 k g$ $5 kg$ and the second with mass $35 k g$ $35 kg$ . If the first weight is $9 m$ $9 m$ from the fulcrum, how far is the second weight from the fulcrum?

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Answer 1 · 2016-12-12T21:22:22

$= 1.29 m (3 s . f .)$ $= 1.29m (3s.f.)$

use turning moments:

$M_{1} \cdot D_{1} = M_{2} \cdot D_{2}$ $M_1 * D_1 = M_2 * D_2$

( $M$ $M$ being mass of weight on lever, $D$ $D$ being distance from fulcrum)

$M_{1} = 5 k g$ $M_1 = 5kg$
$M_{2} = 35 k g$ $M_2 = 35kg$

$D_{1} = 9 m$ $D_1 = 9m$

find $D_{2}$ $D_2$ :

$M_{1} \cdot D_{1} = 5 \cdot 9 = 45$ $M_1 * D_1 = 5 * 9 = 45$
$M_{2} \cdot D_{2} = 35 \cdot D_{2} = 45$ $M_2 * D_2 = 35 * D_2 = 45$

$D_{2} = (\frac{45}{35}) m$ $D_2 = (45/35) m$

$= (\frac{9}{7}) m$ $= (9/7)m$

$= 1.29 m (3 s . f .)$ $= 1.29m (3s.f.)$

A balanced lever has two weights on it, the first with mass 5 kg 5kg5 kg and the second with mass 35 kg35kg35 kg. If the first weight is 9 m9m 9 m from the fulcrum, how far is the second weight from the fulcrum?