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

Published by Patrick Mutisya · 14 days ago

Cambridge A-Level Physics 9702 – Equilibrium of Forces

Equilibrium of Forces

Learning Objective

Understand that the upthrust (buoyant force) acting on an object immersed in a fluid arises from a difference in hydrostatic pressure between the top and bottom surfaces of the object.

Key Concepts

  • Hydrostatic pressure increases with depth: \$p = \rho g h\$

  • Pressure acts perpendicular to any surface in contact with the fluid.

  • Resultant force due to pressure on a submerged surface is the integral of pressure over that surface.

  • Upthrust (buoyant force) \$F_B\$ equals the weight of the displaced fluid.

Derivation of Upthrust from Pressure Difference

Consider a rectangular block of cross‑sectional area \$A\$ and height \$h\$ fully submerged in a fluid of density \$\rho\$.

  1. Pressure at the top surface (depth \$h_1\$):

    \$p{\text{top}} = \rho g h1\$

  2. Pressure at the bottom surface (depth \$h2 = h1 + h\$):

    \$p{\text{bottom}} = \rho g h2 = \rho g (h_1 + h)\$

  3. Force on the top surface (downward):

    \$F{\text{top}} = p{\text{top}} A = \rho g h_1 A\$

  4. Force on the bottom surface (upward):

    \$F{\text{bottom}} = p{\text{bottom}} A = \rho g (h_1 + h) A\$

  5. Net upward force (upthrust):

    \$FB = F{\text{bottom}} - F_{\text{top}} = \rho g h A\$

  6. Since \$V = Ah\$ is the volume of fluid displaced,

    \$F_B = \rho g V\$

    which is exactly the weight of the displaced fluid.

Table: Hydrostatic Pressure at \cdot arious Depths

Depth \$h\$ (m)Pressure \$p = \rho g h\$ (Pa)Force on 0.01 m² area (N)
0.5\$\rho g \times 0.5\$\$\rho g \times 0.5 \times 0.01\$
1.0\$\rho g \times 1.0\$\$\rho g \times 1.0 \times 0.01\$
2.0\$\rho g \times 2.0\$\$\rho g \times 2.0 \times 0.01\$

Application: Conditions for Floating and Sinking

For an object of mass \$m\$ and volume \$V\$ immersed in a fluid:

  • If \$mg < \rho_{\text{fluid}} g V\$, the net force is upward → the object floats.

  • If \$mg = \rho_{\text{fluid}} g V\$, the object is neutrally buoyant → it remains suspended at any depth.

  • If \$mg > \rho_{\text{fluid}} g V\$, the net force is downward → the object sinks.

Common Misconceptions

  • “Buoyancy is a separate force.” In reality, it is the resultant of pressure forces acting on the object's surface.

  • “Only the bottom surface contributes to upthrust.” Both top and bottom surfaces experience pressure; the difference creates the net upward force.

  • “Upthrust depends on the object's weight.” Upthrust depends solely on the volume of fluid displaced, not on the object's own weight.

Suggested Diagram

Suggested diagram: A rectangular block submerged in a fluid showing pressure vectors on the top and bottom faces, depth \$h1\$ and \$h2\$, and the resulting upthrust arrow pointing upward.

Summary

The upthrust on a submerged object is a direct consequence of the hydrostatic pressure gradient in a fluid. By integrating pressure over the object's surface, we find that the net upward force equals the weight of the displaced fluid, providing a clear link between fluid statics and the equilibrium of forces.