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

Question

1603 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-07T06:13:16

The distance is $= 1.15 m$ $=1.15m$

www.thoughtco.com

The mass $M_{1} = 7 k g$ $M_1=7kg$

The mass $M_{2} = 55 k g$ $M_2=55kg$

The distance $a = 9 m$ $a=9m$

Taking moments about the fulcrum

$M_{1} \times a = M_{2} \times b$ $M_1xxa=M_2xxb$

The distance is

$b = \frac{M_{1} \times a}{M_{2}} = \frac{7 \cdot 9}{55} = 1.15 m$ $b=(M_1xxa)/(M_2)=(7*9)/(55)=1.15m$

Answer 2 · 2018-04-07T09:13:13

Approximately $1.15$ $1.15$ meters from the fulcrum

On a balanced lever, we have the following relationship:

$m_{1} d_{1} = m_{2} d_{2}$ $m_1d_1=m_2d_2$

$m_{1}, m_{2}$ $m_1,m_2$ are the masses of the two objects

$d_{1}, d_{2}$ $d_1,d_2$ are the distances of the two objects from the fulcrum

And so, we got:

$7 \ "kg"*9 \ "m"=55 \ "kg"*d_2$

$d_2=(7color(red)cancelcolor(black)"kg"*9 \ "m")/(55color(red)cancelcolor(black)"kg")$

$~~1.15 \ "m"$

A balanced lever has two weights on it, the first with mass 7 kg 7kg7 kg and the second with mass 55 kg55kg55 kg. If the first weight is 9 m9m 9 m from the fulcrum, how far is the second weight from the fulcrum?