Question

Three liquids that will not mix are poured into a cylindrical container. The volumes and densities of the liquids are 0.50 L, 2.6 g/cm³; 0.25 L, 1.0 g/cm³; and 0.40 L, 0.80 g/cm³. What is the force on the bottom of the container due to these liquids? One liter = 1 L = 1000 cm³. (Ignore the contribution due to the atmosphere.)

Answer 1

The force is
`F = 18.33 \ N`

Step-by-step explanation:

From the question we are told that

The number of liquids is n = 3

The volume of the first liquid is
`V_1 = 0.50 L = 0.0005 \ m^3`

The density of the first liquid is
`\rho_1 = 2.6 \ g/cm^3`

The volume of the second liquid is
`V_2 = 0.25 L = 250\ cm^3`

The density of the second liquid is
`\rho_2 = 1.0 \ g/cm^3`

The volume of the third liquid is
`V_3 = 0.40 L = 400\ cm^3`

The density of the third liquid is
`\rho_3 = 0.80 \ g/cm^3`

Generally the force at the bottom of the container is mathematically represented as

`F = m_t * g`

Here
`g = 980.665 \ cm/s^2`

Here
`m_t` is the total mass of all the liquid which is mathematically represented as

`m_t = ( V_1 * \rho_1 )+ ( V_2 * \rho_2)+ ( V_3 * \rho_3)`

=>
`m_t = ( 500 * 2.6)+ ( 250 * 1.0 )+ ( 400 * 0.80 )`

=>
`m_t = 1870 \ g`

So

`F = 1870 * 980.66`

=>
`F = 1833843.55 \ g \cdot cm /s^2`

=>
`F = 1833843.55 \ g \cdot cm /s^2 = (1833843.55)/(1000 * 100) kg \cdot m /s^2`

=>
`F = 18.33 \ N`