Question

Calculate the pressure exerted by a block of L = 1.5m, B = 1m, H = 0.5m and mass 300Kg.

Answer 1

Hint: The pressure is defined as the ratio of perpendicular force to the area of cross section. The SI unit of pressure is Pascal. The weights of the body is the force with which the body is pulled towards the centre of the earth.

Formula used:The formula of the weight is given by,
$ \Rightarrow W = mg$
Where W is the weight the mass of the body is m and the acceleration due to gravity is g.
The formula of the pressure is given by,
$ \Rightarrow P = \dfrac{F}{A}$
Where pressure is P the force is F and the cross sectional area is A.

Complete step by step solution:
It is given in the problem that the length of the block is 1.5m the breath of the block is 1m and the height of the block is 0.5m also the mass of the block is 300 kg.
Let us calculate the weight of the block.
The formula of the weight is given by,
$ \Rightarrow W = mg$
Where W is the weight the mass of the body is m and the acceleration due to gravity is g.
The weight of the block is equal to,
$ \Rightarrow W = mg$
$ \Rightarrow W = 300 \times 10$
$ \Rightarrow W = 3000N$………eq. (1)
The pressure here will be different for different surface areas.
When the surface facing the ground is from length and breath.
Area of cross section is equal to,
$ \Rightarrow {A_1} = 1 \cdot 5 \times 1$
$ \Rightarrow {A_1} = 1 \cdot 5{m^2}$
The pressure for this case is equal to,
$ \Rightarrow P = \dfrac{F}{A}$
$ \Rightarrow {P_1} = \dfrac{{3000}}{{1 \cdot 5}}$
$ \Rightarrow {P_1} = 2000Pa$.
The block faces the ground from the side having length and height.
$ \Rightarrow {A_2} = 1 \cdot 5 \times 0 \cdot 5$
$ \Rightarrow {A_2} = 0 \cdot 75{m^2}$.
The pressure for this case is equal to,
$ \Rightarrow P = \dfrac{F}{A}$
$ \Rightarrow {P_2} = \dfrac{{3000}}{{0 \cdot 75}}$
$ \Rightarrow {P_2} = 4000Pa$.
The surface area of the side having breath and height.
$ \Rightarrow {A_3} = 1 \times 0 \cdot 5$
$ \Rightarrow {A_3} = 0 \cdot 5{m^2}$.
The pressure for this case is equal to,
$ \Rightarrow P = \dfrac{F}{A}$
$ \Rightarrow {P_3} = \dfrac{{3000}}{{0 \cdot 5}}$
$ \Rightarrow {P_3} = 6000Pa$.

The pressure by the block will depend on which side the block is kept on the ground. The pressures are given as${P_1} = 2000Pa$,${P_2} = 4000Pa$ and ${P_3} = 6000Pa$.

Note: It is advisable for the students to understand and remember the formula of the pressure as it is very helpful in solving problems like these. The formula of the weight should also be remembered. The weight in the SI unit is in Newtons.