Buoyancy & Fluid Pressure Simulator

A block with a fixed volume of $V = 0.0010$ m³ (a 10 cm cube) is placed in a tank of fluid. If it floats, it sits at the surface with part of it above the fluid; if it sinks, it rests on the bottom. Arrows show its weight W (down) and the buoyant force F_b (up), with the buoyant arrow drawn in proportion to the weight. The simulator uses $g = 9.8$ m/s².

The panel shows the weight and the buoyant force in newtons, the submerged fraction as a percentage, a verdict (Floats, Neutral or Sinks), the depth h and the gauge pressure in pascals. While the block floats, the weight and buoyant force readouts are always equal. Once the object density passes the fluid density, the block sits fully submerged on the bottom and the buoyant force stops growing: it stays at $\rho_{fluid} V g$ however dense the block becomes.

Try keeping the object density fixed at 800 and raising the fluid density: the block rides higher, because less displaced fluid is needed to hold it up. Then set the depth to 3.0 m and compare the pressure in water (1000) with a denser fluid such as 1250.

Finally, set the two densities equal. The verdict changes to Neutral, the block is 100% submerged, and the buoyant force exactly equals the weight. A neutrally buoyant object, like a fish or a submarine with balanced ballast, stays at whatever depth it is placed.

Density is mass per volume, $\rho = m/V$, so the block's weight is $W = \rho_{obj} V g$. Archimedes' principle says the buoyant force equals the weight of the fluid displaced: $$F_b = \rho_{fluid} V_{disp}\, g$$

A floating object is in equilibrium, so $F_b = W$. Setting $\rho_{fluid} V_{disp} g = \rho_{obj} V g$ gives the submerged fraction $$\frac{V_{disp}}{V} = \frac{\rho_{obj}}{\rho_{fluid}}$$ A sinking object displaces its whole volume, $F_b = \rho_{fluid} V g < W$, and the floor supplies the difference as a normal force: $F_N = W - F_b$.

Pressure in a fluid at rest increases with depth: $$P = P_0 + \rho g h$$ The simulator reports the gauge pressure, $P - P_0 = \rho g h$, which ignores the atmosphere pressing on the surface. Absolute pressure adds about $1.0 \times 10^5$ Pa at sea level.

The buoyant force exists because of this pressure difference: the fluid pushes up on the bottom of the block harder than it pushes down on the top, since the bottom is deeper.

A 10 cm aluminum cube ($\rho = 2700$ kg/m³) is lowered into water. Find its weight, the buoyant force and the force the tank floor exerts on it. Then find the pressure 3.0 m below the surface.

Weight. $W = (2700)(0.0010)(9.8) = 26.46$ N.

Buoyant force. Aluminum is denser than water, so the cube sinks and is fully submerged: $F_b = (1000)(0.0010)(9.8) = 9.80$ N.

Floor force. $F_N = 26.46 - 9.80 = 16.66$ N. If you held the cube on a string instead, this is the tension you would need: it feels about 37% lighter in water.

Pressure. $P_{gauge} = (1000)(9.8)(3.0) = 29{,}400$ Pa, so the absolute pressure is about $1.3 \times 10^5$ Pa. In a denser fluid of 1250 kg/m³, the same depth gives $(1250)(9.8)(3.0) = 36{,}750$ Pa.

In the simulator, set the object density to 2700 and the fluid density to 1000. The panel should show a weight of 26.46 N, a buoyant force of 9.80 N, 100% submerged and Sinks. Set h to 3.0 m to read 29,400 Pa. For a floating case, change the object density to 800: it floats 80% submerged. Raise the fluid density to 1250 and the submerged fraction drops to 64%.

Fluids are Unit 8 of the AP Physics 1 course. Pressure and depth are Topic 8.2, and buoyancy is Topic 8.3, Fluids and Newton's Laws. Questions often ask for a free-body diagram of a floating or submerged object, a comparison of buoyant forces on objects of equal volume, or a prediction of what a scale reads when an object is lowered into water.