# A balanced lever has two weights on it, the first with mass 6 kg  and the second with mass 45 kg. If the first weight is  9 m from the fulcrum, how far is the second weight from the fulcrum?

Jan 19, 2016

$d = 1.2 m$

#### Explanation:

A body which has no tendency to rotate under the combined result of a number of forces acting on it is called to be a balanced state.

The rotational tendency of a force is called Moment of the force.

Moment $=$Force$\times$Distance$= F \times D$

where $D$ is the length of Moment arm, which is perpendicular distance between the line of action of the force and the center of moments.

Also in a balanced lever clockwise moments are equal to clockwise moments.

In the given question, forces acting are two weights for which
$F = m . g$ where $g$ is acceleration due to gravity.

If $d$ is the length of Moment arm, i.e ., perpendicular distance between the weight and the fulcrum, then moment of one force about the fulcrum is equal and opposite to the moment of other force about the fulcrum.

Stating mathematically
$M o m e n {t}_{1} = M o m e n {t}_{2}$

or ${m}_{1.} g . {d}_{1} = {m}_{2.} g . {d}_{2}$
$\implies {m}_{1.} {d}_{1} = {m}_{2.} {d}_{2}$

Inserting given values
$6 \times 9 = 45. {d}_{2}$
or ${d}_{2} = \frac{6 \times 9}{45} m$