Worked Examples-Calculation Part 2 - Pressure

Considered Formula;

$$\text {i)}\quad P = mgh$$ $$\qquad \text {Where}~~P \to Pressure,\quad m \to Mass \quad h \to Height \\\qquad g \to Acceleration~due~to~gravity, \quad h \to Height$$ $$\text {ii)}\quad \frac {F}{A} = \frac {f}{a}$$ $$\qquad \text {Where}~~F \to Force~in~large~piston,\quad A \to Area~of~large~piston \\\qquad f \to Force~in~small~piston,\quad a \to Area~of~small~piston$$

Example 1

A rectangle metal block with sides 1.5m by 1.0m by 1.2m rests on a horizontal surface. If the density of the metal is 7000kg/m³. Calculate the maximum and minimum pressure that the block can exerts on the surface. (Take the weight of 1kg mass to be 10N).

Solution

Data given

Density (ρ)= 7000 Kg/m³

Volume (V) = (1.5 x 1.0 x 1.2) = 1.8 m³

Volume (V) =25 cm³

Accelaration due to gravity (g) =10 m/s²

Mass (m) = ?

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\rho = \frac {m}{V}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~m = \rho V$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 7000 \times 1.8$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 12600~Kg$$

The mass of the metal block = 12600 Kg

Maximum Pressure (P_max) = ?

Minimum area (A_min) = 1.0 x 1.2 m² = 1.2 m²

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~P_{max} = \frac {F}{A_{min}}$$ $$\begin {align}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~P_{max} = \frac {mg}{A_{min}} \quad \Rightarrow Force = Weight = mg \end {align}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac {12600 \times 10}{1.2}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {126000}{1.2}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {126000 \times 10}{1.2 \times 10}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {1260000}{12}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 105000~N/m^2$$

The maximum pressure = 105000 N/m²

Minimum Pressure (P_min) = ?

Maximum area (A_max) = 1.5 x 1.2 m² = 1.8 m²

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~P_{min} = \frac {F}{A_{max}}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~P_{min} = \frac {mg}{A_{max}}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\frac {12600 \times 10}{1.8}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {126000}{1.8}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {126000 \times 10}{1.8 \times 10}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {1260000}{18}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 70000~N/m^2$$

The minimum pressure = 70000 N/m²

Example 2

Calculate the pressure at the bottom of tank of water 15m deep due to the water above it.
(Density of water=1000 kg/m³ and Accelaration due to gravity (g) =10 m/s²)

Solution

Data given

Density (ρ) = 1000 kg/m³

Height (h) = 15 m

Accelaration due to gravity (g) =10 m/s²

Pressure (P) = ?

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~P = \rho gh$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 1000 \times 10 \times 15$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 150000~N/m^2$$

The pressure at the bottom of tank of water = 150000 N/m²

Example 2

Calculate the pressure at the bottom of tank of water 15m deep due to the water above it.
(Density of water=1000 kg/m³ and Accelaration due to gravity (g) =10 m/s²)

Solution

Data given

Density (ρ) = 1000 kg/m³

Height (h) = 15 m

Accelaration due to gravity (g) =10 m/s²

Pressure (P) = ?

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~P = \rho gh$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 1000 \times 10 \times 15$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 150000~N/m^2$$

The pressure at the bottom of tank of water = 150000 N/m²

Example 3

Find the pressure exerted by a force of 240 newtons on an area of 30 cm². Give your answer in newtons/m²

Solution

Data given

$$\text {Area (A)} = 30~cm^2 = \Bigl( \frac {30}{10000}\Bigr)~m = 0.003 m$$

Force (F) = 240 N

Pressure (P) = ?

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~P = \frac {F}{A}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {240}{0.003}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {240 \times 1000}{0.003 \times 1000}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {240000}{3}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 80000~N/m^2$$

The pressure at the bottom of tank of water = 80000 newtons/m²

Example 4

In a hydraulic press the area of the piston to which the effort is applied is 5cm². If the press can raise a weight of 2 KN when an effort of 400 N is applied, what is the area of the piston under the load?

Solution

Data given

Force in large piston (F) = 2 KN = (2 x 1000) N = 2000 N

Force in small piston (f) = 400 N

Area in small piston (a) = 5 cm²N

Area in large piston (A) = ?

Pressure (P) = ?

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\frac {F}{f} = \frac {A}{a}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\frac {2000}{400} = \frac {A}{5}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~A = \frac {2000}{400} \times 5$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 25~m^2$$

The area of the piston under the load = 25 m²