Answer
The force on the bottom $=18.3$ $N$
Work Step by Step
The force on the bottom of the container is due to the total weight of the fluids.
But for finding their weight, we need their mass.
For finding mass, use the relation:
$density =\frac{mass}{volume}$
Rearranging it gives:
$mass=density \times volume$
Therefore the product of density and volume of each liquid will gives their respective masses.
So, for the first liquid :
$mass= 2.6 $ $g/cm^{3} \times 0.50$ $L$
$mass = 1.3$ $\frac{L g}{cm^{3}}$
Here the unit is $\frac{L g}{cm^{3}}$ , but we need the unit in Metric system for carrying the out the further calculations.
For that we use the fact that $1L=1000cm^{3}$ and substitute it in the unit:
$\frac{L g}{cm^{3}} = \frac{ 1000cm^{3}\times g } {cm^{3} }=1000$ $g= 1$ $kg$
So,
mass of first liquid $=1.3$ $kg$
For the second liquid:
$mass=1.0$ $g/cm^{3}\times$ $0.25$ $L=0.25$ $\frac{L g}{cm^{3}} = 0.25$ $kg$
So,
mass of second liquid = $0.25$ $kg$
Similarly for third liquid:
$mass=0.80 g/cm^{3}\times 0.40$ $L = 0.32$ $\frac{L g}{cm^{3}} = 0.32$ $kg$
mass of third liquid $=0.32$ $kg$
So total mass $=1.3$ $kg$ $+0.25$ $kg+0.32$ $kg= 1.87$ $kg$
Using:
$W=mg$
Total weight $= 1.87$ $kg\times 9.8$ $m/s^{2}= 18.32$ $N$
So, the force on the bottom $\approx18.3$ $N$
