Apparent Weight in a Liquid: Subtract Buoyancy from Gravitational Weight

For a submerged object held stationary with upward support and buoyancy, the support reading is T = mg − FB. Archimedes’ principle gives FB = ρfluid g Vdisplaced under the stated static-fluid assumptions.

Draw the three forces

The object has downward gravitational force mg, upward buoyant force FB, and upward support force T in the suspended model used here. Equilibrium gives T + FB = mg. The smaller support reading does not mean the object has lost mass or that gravity has stopped acting.

Use the volume of displaced fluid. For a completely submerged solid object, this is its external submerged volume. A partly immersed object displaces only the volume below the surface. The density in the buoyancy formula belongs to the fluid, not automatically to the object.

Worked support reading

A fictional solid object has mass 2 kg and is fully submerged while suspended. Its external volume is 0.0005 m³. The exercise specifies fluid density 1,000 kg/m³ and g = 10 m/s². Its gravitational weight is 2 × 10 = 20 N.

The buoyant force is 1,000 × 10 × 0.0005 = 5 N. The required support is 20 − 5 = 15 N. Check the balance: support 15 N plus buoyancy 5 N equals the downward weight 20 N.

What changes if it floats freely?

A freely floating object in equilibrium has no suspension support in this simple model. Then FB = mg. Its submerged volume adjusts until it displaces fluid of the same weight as the object. Do not use the fully submerged volume if only part of the object lies below the surface.

If your assumed calculation gives a negative upward support, the assumed suspension arrangement needs reconsideration. A light object held fully underwater may require a downward constraint instead. The sign is information about the force model, not permission to report a negative mass.

Common mistake

Convert cm³ to m³ correctly: 1 cm³ = 10⁻⁶ m³. Volume conversion is cubic. Do not subtract density from weight, because their units differ. This is a paper force-balance exercise; real experimental support readings can also involve other forces, motion or measurement errors.

Mass nahi ghatti; liquid ka upward force support reading ko kam karta hai.

Three practice questions with worked answers

1. Weight is 30 N and buoyant force is 8 N. Find upward support in equilibrium.

   T = 30 − 8 = 22 N. Together support and buoyancy balance weight.

2. Find FB for displaced volume 200 cm³, fluid density 1,000 kg/m³ and g = 10 m/s².

   200 cm³ = 0.0002 m³. Thus FB = 1,000 × 10 × 0.0002 = 2 N.

3. An object floats freely with weight 12 N. What is its buoyant force in equilibrium?

   With no additional vertical support, buoyant force equals weight: FB = 12 N upward.

Related study guides

• Gravitation Class 9 Notes: Universal Law, Value of g, Free Fall and Buoyancy Numericals
• Why Objects Float or Sink: Density Questions Made Easy

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References and authorship

Original examples and practice questions, prepared with AI assistance and editorial arithmetic review. Related definitions and methods: OpenStax: Archimedes’ Principle. No textbook or exam-board affiliation is claimed.