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Pressure is defined as force per unit area. Therefore, pressure = force divided by area. Since force and area are both vectors, how can we perform this division without violating the rule of vectors?

If pressure is a scalar quantity, how did this scalar nature came about from the vector nature of force?

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The pressure is defined by:

$$ \text{d}\mathbf{F}_n = -P\text{d}\mathbf{A} $$

where $\text{d}\mathbf{F}_n$ is the normal force on a surface element $\text{d}\mathbf{A}$. The pressure $P$ is the (scalar) constant of proportionality linking the two. Since the two vectors are in the same direction we can rewrite the equation using the unit vector normal to the surface $\mathbf{n}$:

$$ \text{d}F_n\mathbf{n} = -P\text{d}A\mathbf{n} $$

where $F_n$ and $A$ are now just scalars, and this gives us:

$$ P = \frac{F_n}{A} $$

This last equation only involves scalars.

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  • $\begingroup$ Thanks. It is generally not emphasized that it is the normal component of the force. Perhaps it is understood by those who are well versed. $\endgroup$ – Radhakrishnamurty P Dec 3 '15 at 4:47
  • $\begingroup$ Pressure is not the scalar of a force. This point is of interest to me. $\endgroup$ – Radhakrishnamurty P Dec 3 '15 at 4:48
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Only the force component that's perpendicular (90 degree angle) to the surface will create a pressure. Let's consider a toy car on a track that has a 45 degree incline. Gravity exerts a force on the car. We can split this force into two components: one that is parallel to the track and the other one is perpendicular to the track. Only the perpendicular force will create pressure on the track. The parallel will accelerate the car so it will start moving.

This becomes obvious if you consider the track on a 90 degree incline so it's vertical. The car just falls down. All force is used to accelerate the car and the pressure on the track will be zero.

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