Pressure & Buoyant Forces — Practice · Grade 8 Science

Chapter: Forces

What this practice covers

By the end of Pressure & Buoyant Forces you should be able to:

11 questions, numeric, multiple choice and written answer, each with a worked explanation.

Vocabulary used: pressure, fluid, buoyant force, model, evidence, system boundary, assumption.

Problem 1

A toy floats steadily with weight 3 N. What is its buoyant force?
Answer and explanation

3 N

The upward and downward forces balance.

Problem 2

For a fixed force, spreading it over a larger area makes pressure…
  1. larger
  2. smaller
  3. always zero
  4. independent of area
Answer and explanation

smaller

Pressure = force ÷ area.

Problem 3

Why does a floating object not have “no forces”?
Answer and explanation

Gravity acts downward and buoyancy acts upward; their balance gives zero net force.

Problem 4

A 200 N force is spread over 0.5 m². Find pressure and compare with the same force on 0.25 m². Explain the reasoning in your own words.
Answer and explanation

400 Pa initially; reducing the area by half doubles pressure to 800 Pa.

Pressure=force/area=200/0.5=400 Pa.

Halving area gives 200/0.25=800 Pa.

Response criteria

  • Use the stated conditions.
  • Show the evidence or calculation.
  • Explain the conclusion and any stated limitation.

Problem 5

A floating object weighs 18 N. What is its buoyant force when stationary with no other vertical support? Show the reasoning, not just the conclusion.
Hints

Use displaced fluid volume, not the container's total water volume.

Answer and explanation

18 N upward, balancing its weight.

Problem 6

Two equal-weight objects have different shapes. Can one float and one sink in the same liquid? Explain.
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Yes. Shape can change enclosed volume and average density; a hollow object can displace enough liquid to balance its weight while a compact one cannot.

Problem 7

Which displacement must be smaller in the ideal lift?
Answer and explanation

The larger-area output piston moves a shorter distance when displaced fluid volumes match.

The larger-area output piston moves a shorter distance when displaced fluid volumes match.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 8

Illustrative pistons share fluid pressure 2,000 Pa. Areas are 0.01 and 0.04 m². A floating object separately displaces water with weight 8 N. Calculate piston forces and explain why a force gain is not free energy. Identify the upward force in the floating model.
Answer and explanation

Using force = pressure × area, the piston forces are 20 N and 80 N. In an ideal hydraulic arrangement, the larger force acts through a smaller distance while displaced volumes balance, so energy is not created. The displaced-fluid model gives an 8 N buoyant force; floating equilibrium also requires the object’s weight to be balanced.

Using force = pressure × area, the piston forces are 20 N and 80 N. In an ideal hydraulic arrangement, the larger force acts through a smaller distance while displaced volumes balance, so energy is not created. The displaced-fluid model gives an 8 N buoyant force; floating equilibrium also requires the object’s weight to be balanced.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 9

Deeper fluid exerts greater pressure on an equal-area patchDeeper fluid exerts greater pressure on an equal-area patchStationary waterShallower patchDeeper patchEqual patch areasGreater depth gives greater pressure above the surface reference in this model.Fluid pressure pushes in all directions; only selected side patches are shown.
The selected equal-area side patches compare pressure effects at different depths in stationary water. The deeper patch experiences a greater pressure contribution relative to the surface reference. The arrows are comparative forces on equal-area patches, not a claim that pressure is itself a vector only pointing sideways. A full fluid model includes pushes in all directions and the pressure at the surface.
For this unit’s illustrated case, would a downward-only force drawing fully describe pressure on a submerged object? Explain what is missing.
Answer and explanation

No. Fluid pushes on all surfaces, including sides and top/bottom. Pressure varies with depth; comparing the complete surface forces is needed to explain net buoyancy.

Depth means vertical distance below the fluid surface. Pressure is force per area, and direction belongs to the force on each surface. A single selected arrow cannot represent every force acting on an immersed object.

No. Fluid pushes on all surfaces, including sides and top/bottom. Pressure varies with depth; comparing the complete surface forces is needed to explain net buoyancy.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

Problem 10

Floating, sinking and being held are different casesFloating, sinking and being held are different casesFloating steadilyReleased heavy blockHeld by a stringWeight 3 N downBuoyancy 3 N upNet force = 0 NWeight 4 N downBuoyancy 2 N upNet = 2 N downNet force = 0 NThese are separate cases. Arrows use 25 drawing units per newton.Predict acceleration from the net force; an object can still have velocity.Weight:4 N downBuoyancy:2 N upString:2 N upBlue horizontal lines mark the water surface.
Interpret the different support conditions in this diagram, then apply the same method to the changed force values in the question. The labels are the illustrated examples, not the numerical answer to the new case.
A submerged object is held still by a string. Its weight is 9 N and buoyant force 6 N. What upward string force balances it, and why would drawing only equal buoyancy and weight misdescribe the case?
Answer and explanation

The string supplies 3 N upward. Buoyancy alone is smaller than weight; the third force is needed to make the held object’s net force zero.

List all forces on the object.

Use 6+T−9 = 0 without calling the held state unsupported floating.

The string supplies 3 N upward. Buoyancy alone is smaller than weight; the third force is needed to make the held object’s net force zero.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

Problem 11

Measure displaced volume to predict buoyancyMeasure displaced volume to predict buoyancyBefore: 400 mLFully immersed: 650 mL02004006007000200400600700blockDisplaced water volume = 650 − 400 = 250 mL250 mL = 250 cm³ = 0.00025 m³Displaced water mass = 1,000 × 0.00025 = 0.25 kgBuoyant force = weight of that water = 0.25 × 10 = 2.5 N
The labeled scale records a rise from 400 to 650 mL when the block is fully immersed without bubbles or spilled water. The displaced volume is the change, 250 mL, rather than the final reading. In this model the block is immersed without touching the container floor; an unshown thin holder prevents it moving during measurement. The buoyant force is the weight of 250 mL of water, 2.5 N, independent of whether the block would float or sink when released. For a partly immersed floating object, count only its submerged volume.
In a changed trial, a completely immersed block raises water from 400 to 700 mL. Its mass is 0.50 kg. Use water density 1 g/mL and g = 10 N/kg. Find displaced volume, buoyancy, weight and net force immediately after release from rest. Explain why buoyancy is not zero when it begins to sink.
Answer and explanation

Displaced volume is 300 mL, giving displaced water mass 0.30 kg and buoyancy 3 N up. The block weighs 0.50×10 = 5 N down. With other support absent and initial drag neglected, net force is 2 N down. It begins accelerating downward while water still provides a 3 N upward buoyant force.

700−400 = 300 mL.

300 mL of water has mass 300 g = 0.30 kg, so Fᵦ = 0.30×10 = 3 N.

Object weight = 0.50×10 = 5 N.

5−3 = 2 N down, with buoyancy still acting.

Response criteria

  • Subtract the initial reading and use the fluid density to get buoyancy.
  • Use object mass separately to find weight.
  • Compare forces and distinguish buoyancy from their net result.

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