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?

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Answer 1 · 2018-04-07T06:13:16

The distance is $=1.15m$

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The mass $M_1=7kg$

The mass $M_2=55kg$

The distance $a=9m$

Taking moments about the fulcrum

$M_1xxa=M_2xxb$

The distance is

$b=(M_1xxa)/(M_2)=(7*9)/(55)=1.15m$

Answer 2 · 2018-04-07T09:13:13

Approximately $1.15$ meters from the fulcrum

On a balanced lever, we have the following relationship:

$m_1d_1=m_2d_2$

$m_1,m_2$ 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:

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

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

$~~1.15 \ "m"$

A balanced lever has two weights on it, the first with mass 7 kg 7 kg and the second with mass 55 kg55 kg. If the first weight is 9 m 9 m from the fulcrum, how far is the second weight from the fulcrum?