A spring scale underwater: measure buoyancy and density
You will be able to: Use a force diagram and apparent weight to infer displaced volume and density.
Why does a submerged object pull less on its supporting scale?
A scale reads 40 N when a rigid object hangs in air and 30 N when it hangs fully underwater. Water supports 10 N of the load. The object’s mass and gravitational weight have not decreased.
A useful starting point: Floating is a force balance, not a size contest →
Words and symbols before equations
- Tension T
- Upward force from the supporting scale, in N.
- Apparent weight
- The supporting scale reading; it need not equal gravitational weight.
- Static force balance
- T+B−mg=0 for an object hanging at rest.
- Negligible air buoyancy
- Approximation that the air reading equals mg.
What this picture assumes
Rigid 4 kg object, overall volume 0.001 m³, suspended at rest from above in water (1000 kg/m³), g=10 m/s². No wall/bottom contact; air buoyancy ignored. All shown tensions are positive.
Connect the picture to the physics
Draw three vertical forces on the submerged object: tension and buoyancy upward, weight downward. At rest, T=mg−B. Thus the difference between an air reading and a submerged reading is approximately B if air buoyancy is negligible.
For full immersion, V=(W_air−T_water)/(ρ_f g). Also m=W_air/g, so density can be written ρ_o=ρ_f W_air/(W_air−T_water). Use the same calibrated scale and consistent units; ensure the object is fully wetted, not touching a wall or bottom, and has no trapped bubbles.
A supporting line above can pull up but cannot push down. If the body is less dense than the fluid, it cannot remain fully submerged at rest on a slack line from above alone; it needs a downward restraint. The simple positive-tension setup here uses denser objects. Repeated readings and uncertainties are needed for real measurements.
A worked example, step by step
Air weight is 40 N and underwater tension is 30 N. Use water density 1000 kg/m³ and g=10 m/s². Find volume and average density.
- B=40−30=10 N.
- V=B/(ρ_f g)=10/(1000×10)=0.001 m³.
- Mass=40/10=4 kg.
- ρ_o=m/V=4/0.001=4000 kg/m³. The scale reading changed, not the mass.
Apparent weight is a support force, not a new value of mass. Include buoyancy in the force diagram.
Can the scale reading decrease while true gravitational weight stays the same?
Compare with an explanation
Yes. Buoyancy supplies part of the upward support.
Predict. Change one thing. Explain.
Increase the submerged fraction of a 4 kg, 1 L rigid object in water. Compare weight, buoyancy and supporting tension. Once fully submerged, further depth alone would not change these values in uniform water.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Submerged fraction=1. Weight stays 40 N. B=10 N; scale tension=30 N. 30+10−40=0 N at rest.
Rigid 4 kg object, overall volume 0.001 m³, suspended at rest from above in water (1000 kg/m³), g=10 m/s². No wall/bottom contact; air buoyancy ignored. All shown tensions are positive.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant force, motion or energy relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionAn object reads 30 N in air and 20 N fully submerged in water. Use ρ_f=1000 kg/m³ and g=10 m/s². (a) Find B. (b) Find volume. (c) Find density. (d) State one experimental condition needed for the inference.
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Compare with the answer and four-point rubric
- 1 point: 10 N.
- 1 point: 0.001 m³.
- 1 point: m=3 kg, so ρ_o=3000 kg/m³.
- 1 point: Full immersion without touching walls/bottom or trapping bubbles, with static readings and negligible air buoyancy; any one justified condition earns the point.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What does the underwater scale measure?
Tension, the supporting force.
RECALL 2What does the air-minus-water reading give approximately?
Buoyant force, when air buoyancy is negligible.
RECALL 3Why must the object avoid the bottom?
Otherwise an additional normal force alters the force balance.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
A spring scale underwater: measure buoyancy and density
- Static hanging object: T=mg−B.
- Full immersion: V=(W_air−T_fluid)/(ρ_f g).
- Negligible air buoyancy: ρ_o=ρ_f W_air/(W_air−T_fluid).
Remember: Apparent weight is a support force, not a new value of mass. Include buoyancy in the force diagram.
Conditions: Rigid 4 kg object, overall volume 0.001 m³, suspended at rest from above in water (1000 kg/m³), g=10 m/s². No wall/bottom contact; air buoyancy ignored. All shown tensions are positive.
Refresh Kid · Unit 8 · Objectives 8.3.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 8.3, objectives 8.3.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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