# 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  15 m from the fulcrum, how far is the second weight from the fulcrum?

$45 \setminus m$

#### Explanation:

Let $x$ be the distance of second mass $4$kg from the fulcrum.

Since, the lever is balanced about the fulcrum i.e. net moment about fulcrum must be zero.
Now, taking moment of both the masses about the fulcrum as follows

$12 \setminus \times g \setminus \times 15 = 4 \setminus \times g \setminus \times x$

$x = 45 \setminus m$

Jul 6, 2018

$45$ meters

#### Explanation:

For a balanced lever, we have the relationship:

${m}_{1} {d}_{1} = {m}_{2} {d}_{2}$

where:

• ${m}_{1} , {m}_{2}$ are the masses of the two weights

• ${d}_{1} , {d}_{2}$ are the distances the weights are from the fulcrum

So, we get:

$12 \setminus \text{kg"*15 \ "m"=4 \ "kg} \cdot {d}_{2}$

${d}_{2} = \left(180 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{kg""m")/(4color(red)cancelcolor(black)"kg}}}}\right)$

$= 45 \setminus \text{m}$