The experimental setup should be in equilibrium initially without any load. The setup is given such that $-$ the pivot point is at $40$ mm position $($left side of $0$ mm $),$ load $F_1$ is at $120$ mm position, load $F_2$ is at $40$ mm position $($right side of the pivot point$),$ and load $F_3$ is at $155$ mm position $($far right$).$ There is a mass of $50$ gram already there on far$-$left side $($FI$)$ to make it in equilibrium initially.

$1-$ Record the following data:

$d_1(mm)=,d_3(mm)=195m$

$d_1(m)=\mathrm{.}08mm,d_2(mm)$

Now hang a mass of $220$ gram to the far left $(F_1)$ and $60$ grams to the far right $(F_3),$ and find the masses for load F $2$ that will make the setup back to equilibrium again.

Mass, $m_2(gram)=379$ Mass, $m_2(kg)=-\mathrm{.}937$Load$($ experimental$)F_2($ Newton $)=$

$2-$ Apply equilibrium of moment equation ${\displaystyle \underset{?}{\overset{?}{}}}{M}_0=0$ about the pivot point and use $g=9\mathrm{.}81\frac{m}{s^2}$ to calculate: Mass, grams