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

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

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.