# A balanced lever has two weights on it, the first with mass 7 kg  and the second with mass 55 kg. If the first weight is  9 m from the fulcrum, how far is the second weight from the fulcrum?

Apr 7, 2018

The distance is $= 1.15 m$

#### Explanation:

The mass ${M}_{1} = 7 k g$

The mass ${M}_{2} = 55 k g$

The distance $a = 9 m$

${M}_{1} \times a = {M}_{2} \times b$

The distance is

$b = \frac{{M}_{1} \times a}{{M}_{2}} = \frac{7 \cdot 9}{55} = 1.15 m$

Apr 7, 2018

Approximately $1.15$ meters from the fulcrum

#### Explanation:

On a balanced lever, we have the following relationship:

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

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

• ${d}_{1} , {d}_{2}$ are the distances of the two objects from the fulcrum

And so, we got:

$7 \setminus \text{kg"*9 \ "m"=55 \ "kg} \cdot {d}_{2}$

${d}_{2} = \left(7 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{kg"*9 \ "m")/(55color(red)cancelcolor(black)"kg}}}}\right)$

$\approx 1.15 \setminus \text{m}$