Question

A brick weighing 20N and having dimensions 20cm×10cm×5cm is kept on the ground in three different ways.Calculate the pressure in all three cases.When is the pressure greatest?When it is least?

Answer 1

Given:

Weight of brick, F= 20 N

Dimensions of brick= 20cm×10cm×5cm

To Find:

Pressure in the all the cases in which brick is kept

Solution:

We know that,

• Pressure P exerting on certain surface is given by

$\pink{\boxed{\bf{P=\frac{Force\ or\ Weight}{Area}}}}$

$\rule{190}{1}$

Case-1: When dimension of surface in contact is 20cm×10cm

Let the area of the surface be A₁

So,

$\longrightarrow\rm{A_{1}=20\times10=200\ cm^{2}=0.02\ m^{2}}$

Let the pressure on the given surface be P ₁

So,

$\longrightarrow\rm{P_{1}=\dfrac{F}{A_{1}}}$

$\longrightarrow\rm{P_{1}=\dfrac{20}{0.02}}$

$\longrightarrow\rm\green{P_{1}=1000\ Pa}$

$\rule{190}{1}$

Case-2: When dimension of surface in contact is 20cm×5cm

Let the area of the surface be A₂

So,

$\longrightarrow\rm{A_{2}=20\times5=100\ cm^{2}=0.01\ m^{2}}$

Let the pressure on the given surface be P ₂

So,

$\longrightarrow\rm{P_{2}=\dfrac{F}{A_{2}}}$

$\longrightarrow\rm{P_{2}=\dfrac{20}{0.01}}$

$\longrightarrow\rm\green{P_{2}=2000\ Pa}$

$\rule{190}{1}$

Case-3: When dimension of surface in contact is 10cm×5cm

Let the area of the surface be A₃

So,

$\longrightarrow\rm{A_{3}=10\times5=50\ cm^{2}=0.005\ m^{2}}$

Let the pressure on the given surface be P ₃

So,

$\longrightarrow\rm{P_{3}=\dfrac{F}{A_{3}}}$

$\longrightarrow\rm{P_{3}=\dfrac{20}{0.005}}$

$\longrightarrow\rm\green{P_{3}=4000\ Pa}$

$\rule{190}{1}$

So, pressure in the third case is maximum where dimensions of contact surface is 10cm×5cm

And pressure in the first case is minimum where dimensions of contact surface is 20cm×10cm