A balanced lever has two weights on it, the first with mass 12 kg  and the second with mass 4 kg. If the first weight is  7 m from the fulcrum, how far is the second weight from the fulcrum?

Jan 14, 2018

The distance is $= 21 m$

Explanation:

The mass ${M}_{1} = 12 k g$

The mass ${M}_{2} = 4 k g$

The distance $a = 7 m$

Taking moments about the fulcrum

$12 g \cdot 7 = 4 g \cdot b$

$b = \frac{12 g \cdot 7}{4 g} = 21 m$

Jan 14, 2018

$21.1$m

Explanation:

the lever is balanced, meaning that the moments on either side are equal.

the moment is the force of the weight multiplied by the distance of the weight from the fulcrum.
moment (Nm) $= F \cdot d$

the moment of the weight with mass $12$kg and distance $7$m is equal to the moment of the weight with mass $4$kg and $d$m from the fulcrum.

note: $W = m g$
weight (in newtons) = mass * gravitational field strength (~9.81)

mass $12$kg = weight $118$N (3s.f.)
mass $4$kg = weight $39.2$N (3s.f.)

$118$N $\cdot 7$m $= 826$Nm
$39.2$N $\cdot d$m $= 826$Nm

$d = \frac{826}{39.2} = 21.1$

$d = 21.1$m (3s.f.)