Question

A rectangular solid block has sides $4\times 10\times 20cm$and a density of $7.5gc{m}^{-3}$. If it rests on a horizontal flat surface, calculate the minimum and maximum pressure it can exert (in SI units). Also, calculate the thrust in both cases.

Answer 1

Step 1: Given data

$$\begin{gathered} Sides=4\times 10\times 20cm \\ Density\left(\rho \right)=7.5gc{m}^{-3} \end{gathered}$$

Step 2: Calculating the thrust

The thrust will be equal to the weight of the block.

$$\begin{gathered} Thrust=Weight=m\times g=\rho \times V\times g \\ Volume\left(V\right)=4cm\times 10cm\times 20cm \\ =800c{m}^3 \\ Thrust=7.5gc{m}^{-3}\times 800c{m}^3\times 9.8m{s}^{-2} \\ =60073.5gm{s}^{-2}=\frac{60073.5}{1000}kgm{s}^{-2} \\ =60kgm{s}^{-2}=60N \end{gathered}$$

Step 3: Calculating the maximum pressure

$Pressure=\frac{Thrust}{Area}$

Pressure will be maximum when the block is resting with the face having the minimum area $(4cm\times 10cm)$ downwards.

$$\begin{gathered} P_{max}=\frac{Thrust}{Are{a}_{min}} \\ =\frac{60N}{4cm\times 10cm}=\frac{60N}{{\displaystyle \frac{4}{100}}m\times {\displaystyle \frac{10}{100}}m} \\ =\frac{60\times 100\times 100}{4\times 10} \\ =15000Pa \end{gathered}$$

Step 4: Calculating the minimum pressure

$Pressure=\frac{Thrust}{Area}$

Pressure will be minimum when the block is resting with the face having the maximum area $(20cm\times 10cm)$ downwards.

$$\begin{gathered} P_{min}=\frac{Thrust}{Are{a}_{max}} \\ =\frac{60N}{20cm\times 10cm}=\frac{60N}{{\displaystyle \frac{20}{100}m\times \frac{10}{100}m}} \\ =\frac{60\times 100\times 100}{20\times 10} \\ =3000Pa \end{gathered}$$

Hence,

1. Thrust = $60N$
2. Maximum pressure = $15000Pa$
3. Minimum pressure = $3000Pa$