# A balanced lever has two weights on it, the first with mass 9 kg  and the second with mass 32 kg. If the first weight is  6 m from the fulcrum, how far is the second weight from the fulcrum?

##### 2 Answers
Feb 25, 2016

$\frac{27}{16}$ m.

#### Explanation:

For a balanced lever the torque on the two ends must be equal.

Torque can be defined as $\text{mass" xx "distance from fulcrum}$

So
$\textcolor{w h i t e}{\text{XXX}} {m}_{1} \times {d}_{1} = {m}_{2} \times {d}_{2}$

or for the given data:
color(white)("XXX")9 " kg"xx6" m" = 32" kg" xx d_2 " m"

color(white)("XXX")d_2=(9xx6)/32 " m" = 27/16" m"

Feb 25, 2016

Second weight is $\frac{54}{32} = 1.6875$ $m$ far from fulcrum.

#### Explanation:

Let the the second weight be $x$ m far from the fulcrum.

As the lever is balanced, moment ($M = F \cdot d$, where $F$ is force applied and $d$ is its distance from fulcrum) of the two weights should be equal.

Moment of first weight is $9 \cdot 6 = 54$ and as it balances second weight of $32$ kg, the latter's distance should be

$\frac{54}{32} = 1.6875$ $m$.