Question

A hammer exerts a force of 1.5N on each of the two nails A and B. The area of cross-section of the tip of nail A is $2{\text{m}}{{\text{m}}^2}$ while that of nail B is ${\text{6m}}{{\text{m}}^2}$. Calculate the pressure on each nail.

Answer 1

The pressure on an object refers to the force exerted on the object per unit area. Here the exerted force is the same for both the nails A and B. However, the cross-sectional area of the tip of the nail is different. So more pressure will be for that nail with the lesser cross-sectional area i.e. nail A will have to experience more pressure.
Formula used:
The pressure exerted on an object is given by, $P = \dfrac{F}{A}$ where $F$ is the applied force and $A$ is the area of the object on which the force is applied.

Complete step by step answer:
The force applied on nail A and nail B is the same and it is given to be ${F_A} = {F_B} = 1.5{\text{N}}$.
The cross-sectional area of the tip of nail A is given to be ${A_A} = 2{\text{m}}{{\text{m}}^2}$.
The cross-sectional area of the tip of nail B is given to be ${A_B} = 6{\text{m}}{{\text{m}}^2}$.
We are to find the pressure ${P_A}$ on nail A and the pressure ${P_B}$ on nail B.
Expressing the relation for the pressure on both the nails A and B.
The pressure on nail A is expressed as ${P_A} = \dfrac{{{F_A}}}{{{A_A}}}$ ------- (1)
Substituting for ${F_A} = 1.5{\text{N}}$ and ${A_A} = 2 \times {10^{ - 6}}{{\text{m}}^2}$ in equation (1) we get,
$\Rightarrow {P_A} = \dfrac{{1.5}}{{2 \times {{10}^{ - 6}}}} = 7.5 \times {10^5}{\text{Pa}}$
Thus the pressure on nail A is ${P_A} = 7.5 \times {10^5}{\text{Pa}}$.
Now the pressure on nail B is expressed as
${P_B} = \dfrac{{{F_B}}}{{{A_B}}}$ ------- (2)
Substituting for ${F_B} = 1.5{\text{N}}$ and ${A_B} = 6 \times {10^{ - 6}}{{\text{m}}^2}$ in equation (2) we get,
${P_B} = \dfrac{{1.5}}{{6 \times {{10}^{ - 6}}}} = 2.5 \times {10^5}{\text{Pa}}$

Therefore, the pressure on nail B is ${P_B} = 2.5 \times {10^5}{\text{Pa}}$.

Note:
While substituting the values of various physical quantities in any equation, make sure that all the physical quantities are expressed in their respective S.I units. If this is not the case then the necessary conversion of units must be done. Here the cross-sectional areas of the two nails were expressed in millimetres as ${A_A} = 2{\text{m}}{{\text{m}}^2}$ and ${A_B} = 6{\text{m}}{{\text{m}}^2}$. So before substituting their areas in equations (1) and (2), the values were converted into meters as ${A_A} = 2 \times {10^{ - 6}}{{\text{m}}^2}$ and ${A_B} = 6 \times {10^{ - 6}}{{\text{m}}^2}$.