use the equation ∆p = ρg∆h

Equilibrium of Forces – Cambridge International AS & A Level Physics 9702

Learning Objectives

By the end of this lesson you will be able to:

• Define density, pressure (absolute and gauge), atmospheric pressure and hydrostatic pressure.
• Derive and apply the hydrostatic‑pressure relationship \(\Delta p = \rho g \Delta h\).
• Distinguish between absolute and gauge pressure and use \(p = p_{0}+\rho g h\) where required.
• State and use Archimedes’ principle to find buoyant forces.
• Apply the two static‑equilibrium conditions:
  • \(\displaystyle\sum\vec F = \mathbf 0\) (force equilibrium)
  • \(\displaystyle\sum M = 0\) (moment equilibrium) about any axis.
• Explain moments, couples, torque and adopt a clear sign convention (anticlockwise = positive, clockwise = negative).
• Construct a vector (force) triangle for a coplanar three‑force system.
• Calculate the resultant hydrostatic force on a submerged plane surface and locate its centre of pressure.
• Interpret the results to assess design safety for hinges, supports and brackets.

Fundamental Definitions

Symbol	Quantity	Definition	Units
\(\rho\)	Density	\(\displaystyle \rho = \frac{m}{V}\)	kg m⁻³
g	Acceleration due to gravity	Standard value ≈ 9.81 m s⁻²	m s⁻²
h	Depth below the free surface	Vertical distance measured downwards from the fluid surface	m
p	Absolute pressure	\(\displaystyle p = p_{0} + \rho g h\)	Pa (N m⁻²)
p_{0}	Atmospheric (surface) pressure	Pressure exerted by the surrounding air at the free surface (≈ 1.01 × 10⁵ Pa at sea level)	Pa
\(\Delta p\)	Gauge (pressure difference)	\(\displaystyle \Delta p = p - p_{0}= \rho g h\)	Pa
F	Resultant force on a surface	Integral of pressure over the area	N
M	Moment (torque)	\(\displaystyle M = F\,d\) where \(d\) is the perpendicular distance from the line of action of the force to the chosen pivot	N m

Hydrostatic Pressure – Derivation

Consider a vertical column of fluid of cross‑sectional area \(A\) and height \(\Delta h\).

1. Weight of the column: \(\displaystyle W = \rho\,A\,\Delta h\,g\).
2. The pressure at the bottom exceeds that at the top by \(\Delta p\); the net upward force due to this pressure difference is \(\Delta p\,A\).

For static equilibrium of the fluid column:

\[ \Delta p\,A = \rho\,A\,\Delta h\,g \;\Longrightarrow\; \boxed{\Delta p = \rho g \Delta h} \]

Absolute vs. Gauge Pressure

• Gauge pressure is the pressure above atmospheric: \(\Delta p = \rho g h\).
• Absolute pressure includes the atmospheric contribution: \(p = p_{0} + \rho g h\).
• When only pressure differences are required (e.g. force on a submerged surface) the atmospheric term cancels and may be omitted.

Archimedes’ Principle

The upward buoyant force on an object wholly or partially immersed in a fluid equals the weight of the fluid displaced:

\[ F_{\text{buoyancy}} = \rho_{\text{fluid}}\,g\,V_{\text{disp}} \]

\(V_{\text{disp}}\) is the volume of fluid displaced by the object.

Static Equilibrium of Forces

A rigid body is in static equilibrium when both of the following conditions are satisfied:

1. Force equilibrium: \(\displaystyle\sum\vec F = \mathbf 0\).
2. Moment equilibrium: \(\displaystyle\sum M = 0\) about any chosen axis.

Moments, Couples and Torque – Sign Convention

• Moment (torque): \(M = F d\) (units N m).
• Sign convention: Anticlockwise moments are taken as **positive**; clockwise moments are **negative**. This convention must be applied consistently throughout a problem.
• Couple: Two equal and opposite forces whose lines of action do not coincide. The resulting moment is \(M_{\text{couple}} = F d\) where \(d\) is the perpendicular separation of the forces.
• Principle of moments: In equilibrium the algebraic sum of all clockwise moments equals the sum of all anticlockwise moments.

Worked Example – Moment on a Hinged Gate

A rectangular gate 2.0 m high, 1.5 m wide is hinged at its top edge. The top edge lies 0.5 m below the water surface.

1. Depth at the top: \(h_{1}=0.5\;\text{m}\); at the bottom: \(h_{2}=h_{1}+h=2.5\;\text{m}\).
2. Pressures: \[ p_{1}= \rho g h_{1}=1000\times9.81\times0.5=4.91\times10^{3}\;\text{Pa} \] \[ p_{2}= \rho g h_{2}=1000\times9.81\times2.5=2.45\times10^{4}\;\text{Pa} \]
3. Average pressure: \(\displaystyle\bar p=\frac{p_{1}+p_{2}}{2}=1.47\times10^{4}\;\text{Pa}\).
4. Area of the gate: \(A = h\,b = 2.0\times1.5 = 3.0\;\text{m}^{2}\).
5. Resultant hydrostatic force: \[ F = \bar p\,A = 1.47\times10^{4}\times3.0 = 4.41\times10^{4}\;\text{N} \]
6. Centre of pressure for a vertical plate: \[ y_{cp}= \frac{h_{1}+ \tfrac{h}{3}}{2}= \frac{0.5+ \tfrac{2.0}{3}}{2}=1.17\;\text{m} \] measured from the free surface, i.e. \(0.67\;\text{m}\) below the hinge.
7. Moment about the hinge (anticlockwise = positive): \[ M = F \times 0.67 = 4.41\times10^{4}\times0.67 = 2.95\times10^{4}\;\text{N m}\;(+) . \]

Vector (Force) Triangle – Three‑Force Equilibrium

If three coplanar forces act on a body and the body is in equilibrium, the forces can be placed head‑to‑tail to form a closed triangle.

Example – Sign‑Supported Sign

• Weight \(W = 200\;\text{N}\) (downwards).
• Tension \(T_{1}=150\;\text{N}\) at \(30^{\circ}\) above the horizontal (right‑hand side).
• Find the magnitude and direction of the third tension \(T_{2}\) (left side).

1. Draw \(W\) vertically downwards.
2. From the head of \(W\) draw \(T_{1}\) at \(30^{\circ}\) to the horizontal.
3. The closing side of the triangle represents \(T_{2}\). Measure its length and angle.
4. Using the law of sines: \[ \frac{T_{2}}{\sin30^{\circ}} = \frac{150}{\sin\theta}\quad\Rightarrow\quad T_{2}\approx 260\;\text{N} \] where \(\theta\) is the angle opposite \(W\). The direction of \(T_{2}\) is \(45^{\circ}\) above the horizontal, acting to the left.

Force on Submerged Plane Surfaces

Pressure at a Depth

\[ p = p_{0} + \rho g h \qquad\text{(absolute pressure)} \]

When only the pressure difference is required, use \(\Delta p = \rho g h\).

Resultant Hydrostatic Force

For any plane surface:

\[ F = \int_{A} p\,\mathrm dA \]

When pressure varies linearly (most practical cases) the average‑pressure method is valid:

\[ F = \bar p\,A = \frac{p_{\text{top}}+p_{\text{bottom}}}{2}\;A \]

Centre of Pressure

• Vertical plate (top edge at depth \(h_{1}\), height \(h\)): \[ y_{cp}= \frac{h_{1}+ \tfrac{h}{3}}{2} \] measured from the free surface.
• Inclined plate (inclination \(\theta\) to the vertical): \[ y_{cp}= \bar h + \frac{I_{G}}{\bar p\,A} \] where
  • \(\bar h\) = depth of the centroid,
  • \(I_{G}\) = second moment of area about the horizontal axis through the centroid,
  • \(\bar p = p_{0} + \rho g \bar h\).

Worked Example – Circular Hatch at the Bottom of an Oil Tank

A cylindrical tank of radius 1.2 m is filled with oil (\(\rho = 850\;\text{kg m}^{-3}\)) to a depth of 3.5 m. A circular hatch of diameter 0.5 m is located at the bottom.

1. Pressure at the bottom (gauge): \[ \Delta p = \rho g h = 850 \times 9.81 \times 3.5 = 2.92\times10^{4}\;\text{Pa} \]
2. Area of the hatch: \[ A = \pi\left(\frac{0.5}{2}\right)^{2}= \pi(0.25)^{2}= 1.96\times10^{-1}\;\text{m}^{2} \]
3. Resultant force on the hatch: \[ F = \Delta p \, A = 2.92\times10^{4}\times1.96\times10^{-1}=5.73\times10^{3}\;\text{N} \]

Worked Example – Inclined Plate Hinged at Its Lower Edge

A rectangular plate \(1.0\;\text{m} \times 0.5\;\text{m}\) is hinged at its lower edge and inclined \(30^{\circ}\) to the vertical. The top edge is 0.8 m below the water surface.

1. Depth of the centroid: \[ \bar h = 0.8\;\text{m} + \frac{0.5}{2}\sin30^{\circ}=0.8+0.125=0.925\;\text{m} \]
2. Pressures at the top and bottom: \[ p_{\text{top}}=\rho g (0.8)=1000\times9.81\times0.8=7.85\times10^{3}\;\text{Pa} \] \[ p_{\text{bottom}}=\rho g (0.8+0.5\sin30^{\circ})=1000\times9.81\times0.925=9.07\times10^{3}\;\text{Pa} \]
3. Average pressure: \[ \bar p = \frac{p_{\text{top}}+p_{\text{bottom}}}{2}=8.46\times10^{3}\;\text{Pa} \]
4. Area of the plate: \(A = 1.0\times0.5 = 0.50\;\text{m}^{2}\).
5. Resultant hydrostatic force: \[ F = \bar p\,A = 8.46\times10^{3}\times0.50 = 4.23\times10^{3}\;\text{N} \]
6. Second moment of area about the centroidal horizontal axis: \[ I_{G}= \frac{b h^{3}}{12}= \frac{1.0\,(0.5)^{3}}{12}=5.21\times10^{-3}\;\text{m}^{4} \]
7. Centre of pressure depth: \[ y_{cp}= \bar h + \frac{I_{G}}{\bar p\,A}=0.925+\frac{5.21\times10^{-3}}{8.46\times10^{3}\times0.50}=0.925+1.23\times10^{-3}\approx0.926\;\text{m} \]
8. Perpendicular distance from the hinge to the line of action (≈ 0.926 m). Torque about the hinge (anticlockwise positive): \[ M = F \times 0.926 = 4.23\times10^{3}\times0.926 = 3.92\times10^{3}\;\text{N m}\;(+) \]

Practice Questions

1. Derive the expression for the centre of pressure on a plane surface inclined at an angle \(\theta\) to the vertical, starting from the definition \(y_{cp}= \bar h + I_{G}/(\bar p A)\).
2. A cylindrical tank of radius 1.2 m is filled with oil (\(\rho = 850\;\text{kg m}^{-3}\)) to a depth of 3.5 m.
   • Calculate the absolute pressure at the bottom of the tank (include atmospheric pressure of \(1.01\times10^{5}\) Pa).
   • Determine the total force on a circular hatch of diameter 0.5 m located at the bottom.
3. Explain why a fluid at rest cannot support shear stress and relate this to the condition \(\sum\vec F = \mathbf 0\) for static equilibrium.
4. Three forces act on a bracket in the horizontal plane:
   • Force \(A = 120\;\text{N}\) acting east.
   • Force \(B = 180\;\text{N}\) acting at \(45^{\circ}\) north of east.
   • Force \(C\) of unknown magnitude acting due north.
   Using a vector triangle, find the magnitude of \(C\) and state whether the system is in equilibrium.
5. A rectangular plate (1 m × 0.5 m) is hinged at its lower edge and inclined at \(30^{\circ}\) to the vertical. The top edge is 0.8 m below the water surface. Determine the resultant hydrostatic force and the torque about the hinge (show all steps).

Summary Table of Key Variables

Symbol	Quantity	Units	Typical Value (Water)
\(\rho\)	Density of fluid	kg m⁻³	1000
g	Acceleration due to gravity	m s⁻²	9.81
h	Depth below free surface	m	–
p_{0}	Atmospheric pressure	Pa	1.01 × 10⁵
\(\Delta p\)	Gauge pressure (pressure difference)	Pa	\(\rho g h\)
p	Absolute pressure	Pa	\(p_{0}+\rho g h\)
F	Resultant hydrostatic force	N	–
M	Moment (torque)	N m	–

Key Take‑aways

• Static equilibrium requires both \(\sum\vec F = \mathbf 0\) and \(\sum M = 0\) (with a consistent sign convention).
• Pressure in a fluid at rest increases linearly with depth: \(\Delta p = \rho g \Delta h\). Add \(p_{0}\) for absolute pressure.
• Gauge pressure is sufficient when only pressure differences matter (e.g., force on a submerged surface).
• Archimedes’ principle provides a quick way to find buoyant forces: weight of displaced fluid.
• Moments are calculated as \(M = F d\); anticlockwise = positive, clockwise = negative.
• Resultant hydrostatic force on a plane surface can be obtained using the average‑pressure method; the centre of pressure lies deeper than the centroid.
• Vector (force) triangles give a graphical solution for three‑force equilibrium problems.
• Accurate location of the centre of pressure is essential for the safe design of hinges, brackets and other supports.