understand that the upthrust acting on an object in a fluid is due to a difference in hydrostatic pressure

Equilibrium of Forces

Objective: Understand that the upthrust acting on an object in a fluid is due to a difference in hydrostatic pressure. 🚤

1. Hydrostatic Pressure Basics

Hydrostatic pressure increases with depth because more fluid is above a given point. The pressure at a depth \$h\$ in a fluid of density \$\rho\$ is given by:

\$P = \rho g h\$

  • \$\rho\$ – density of the fluid (kg m⁻³)

  • \$g\$ – acceleration due to gravity (≈9.81 m s⁻²)

  • \$h\$ – depth below the surface (m)

2. Pressure Difference Around a Submerged Object

Imagine a small ball submerged in water. The water pushes on every side of the ball, but the pressure is higher at the bottom than at the top because the bottom is deeper. This creates a net upward force.

  1. Pressure at the top: \$P{\text{top}} = \rho g h{\text{top}}\$

  2. Pressure at the bottom: \$P{\text{bottom}} = \rho g h{\text{bottom}}\$

  3. Net upward pressure difference: \$\Delta P = P{\text{bottom}} - P{\text{top}} = \rho g (h{\text{bottom}} - h{\text{top}})\$

The area over which this pressure acts is the cross‑sectional area \$A\$ of the ball, so the upward force (upthrust) is:

\$F_B = \Delta P \, A = \rho g V\$

where \$V\$ is the volume of the ball. Notice that \$F_B\$ depends only on the volume of fluid displaced, not on the shape of the object. This is Archimedes' principle.

3. Equilibrium of Forces in a Fluid

An object floats when the upthrust equals its weight:

\$\rho{\text{fluid}} V{\text{displaced}} g = m_{\text{object}} g\$

Simplifying, the object will float if its density \$\rho{\text{object}}\$ is less than the fluid's density \$\rho{\text{fluid}}\$.

4. Analogy: The Balloon in Water

Think of a helium balloon in air. The air pushes on the balloon from all sides, but because the balloon is lighter than the air, it rises. Similarly, an object in a fluid experiences a net upward push if it is lighter than the fluid. The difference in pressure at different depths is the “push” that keeps the balloon (or the object) afloat. 🎈

5. Example Problem

A wooden block with a volume of \$0.02\,\text{m}^3\$ and a density of \$600\,\text{kg m}^{-3}\$ is placed in water (\$\rho_{\text{water}} = 1000\,\text{kg m}^{-3}\$). Will it float or sink?

  1. Calculate the block’s mass: \$m = \rho_{\text{block}} V = 600 \times 0.02 = 12\,\text{kg}\$.

  2. Weight: \$W = mg = 12 \times 9.81 \approx 117.7\,\text{N}\$.

  3. Upthrust: \$FB = \rho{\text{water}} V g = 1000 \times 0.02 \times 9.81 \approx 196.2\,\text{N}\$.

  4. Since \$F_B > W\$, the block will float.

6. Examination Tips

TipWhy It Helps
Remember \$F_B = \rho V g\$ – no need to calculate pressure differences for every problem.Quick calculation saves time during exams.
Check units carefully: kg, m, s, N.Unit consistency prevents calculation errors.
Use diagrams to show pressure acting on different parts of the object.Visuals help you explain equilibrium clearly.

Remember: Upthrust comes from the pressure difference between the top and bottom of an object in a fluid. If the upthrust equals the weight, the object is in equilibrium and will float. ⚖️