---
title: "Pascal's law"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:pascals-law"
source: https://www.bctengineer.com/q/semester_6/hydro_power/pascals-law
site: BCT Engineering
---

# Pascal's law

Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics

- Asked in: 2080 Asoj, 5 marks
- Video: [Pascal’s Law](https://www.youtube.com/watch?v=M33EathKb00) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/pascals-law
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

State and prove Pascal's law.

## Short answer

Pascal's law: pressure at a point in a static fluid is equal in all directions. Proof: on a small wedge element, horizontal and vertical force balances yield pₓ = p_z and p_y = p_z as size → 0. Applied in hydraulic presses and jacks.

## Answer

**Pascal's law:** the pressure at a point in a fluid at rest is the same in all directions, i.e. $p_x = p_y = p_z$.

#### Proof
Consider a small wedge-shaped fluid element at rest, of unit width perpendicular to the paper, with sides $dx$ (horizontal), $dy$ (vertical) and inclined face $ds$ at angle $\theta$ to the vertical.

```text
          |\
          | \  ds  (pressure p_z
   p_x -> |  \       normal to it)
       dy |   \
          |    \
          |_____\
            dx
            ^ p_y
```
Pressures: $p_x$ on the vertical face, $p_y$ on the bottom face, $p_z$ on the inclined face. Geometry: $dy = ds\cos\theta$, $dx = ds\sin\theta$ (with $\theta$ the angle between the inclined face and the vertical).

Since the fluid is at rest, there are no shear forces; only pressure forces and weight act.

**Horizontal equilibrium:**
$$
p_x (dy \cdot 1) - p_z (ds \cdot 1)\cos\theta = 0
$$
Using $ds\cos\theta = dy$: $p_x\,dy = p_z\,dy \Rightarrow p_x = p_z$.

**Vertical equilibrium:**
$$
p_y (dx\cdot 1) - p_z (ds\cdot 1)\sin\theta - \rho g \frac{dx\, dy}{2}\cdot 1 = 0
$$
Using $ds\sin\theta = dx$:
$$
p_y\,dx - p_z\,dx - \rho g \frac{dx\,dy}{2} = 0
$$
As the element shrinks to a point ($dy \to 0$), the weight term (second-order small) vanishes, so $p_y = p_z$.

Therefore
$$
p_x = p_y = p_z
$$
Since $\theta$ was arbitrary, the pressure at a point is the same in every direction.

#### Applications
Hydraulic press, hydraulic jack and brakes, where a small force on a small piston gives a large force on a large piston ($F_2 = F_1 A_2/A_1$).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Pascal's law states that in a fluid at rest, the pressure at a point:
   A. Increases with depth only
   B. Is the same in all directions
   C. Acts only vertically
   D. Is zero on the surface
   Answer: B. p_x = p_y = p_z at a point in a static fluid. (Key point: Statement)

2. The proof uses a small wedge element at rest. Why are no shear forces included?
   A. Shear forces are negligible for small wedges
   B. A fluid at rest cannot sustain shear, so only pressure forces and weight act
   C. The wedge is made of solid
   D. Shear forces cancel with weight
   Answer: B. Static fluid has no shear; equilibrium involves only normal pressure forces and weight. (Key point: Proof set-up)

3. Why does p_y = p_z follow in the vertical equilibrium?
   A. The weight term is second-order small and vanishes as the element shrinks to a point
   B. Weight is exactly cancelled by p_x
   C. Weight is zero for any fluid
   D. dx tends to infinity
   Answer: A. The weight (rho g dx dy/2) involves two small lengths and vanishes as dy tends to 0, leaving p_y dx = p_z dx. (Key point: Vertical equilibrium)

4. In a hydraulic press, force on the large piston is:
   A. F2 = F1 A1/A2
   B. F2 = F1 A2/A1
   C. F2 = F1
   D. F2 = F1 (A2 - A1)
   Answer: B. Equal pressure on both pistons gives F2/A2 = F1/A1. (Key point: Application)

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