A balanced lever has two weights on it, the first with mass $15 k g$ $15 kg$ and the second with mass $9 k g$ $9 kg$ . If the first weight is $4 m$ $4 m$ from the fulcrum, how far is the second weight from the fulcrum?

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Answer 1 · 2016-12-09T14:49:34

The distance is $= 6.67 m$ $=6.67m$

The moments on both sides of the fulcrum must de equal on equilibrium.

$M o m e n t =$ $Moment =$ Force x distance

So,

Moment of first weight $= 15 g \cdot 4 = 60 g N m$ $=15g*4=60g Nm$

Moment of the second weight $= 9 g \cdot d N m$ $=9g*d Nm$

Where $d =$ $d=$ distance from the fulcum

On equilibrium

$60 g = 9 g \cdot d$ $60g=9g*d$

$d = \frac{60}{9} = \frac{20}{3} = 6.67 m$ $d=60/9=20/3=6.67m$

A balanced lever has two weights on it, the first with mass 15kg15 kg and the second with mass 9kg9 kg. If the first weight is 4m 4 m from the fulcrum, how far is the second weight from the fulcrum?