Physlets run in a Java-enabled browser on the latest Windows & Mac operating systems.
If Physlets do not run, click here for help on updating Java and setting Java security.
Illustration 14.3: Buoyant Force
Please wait for the animation to completely load.
The buoyant force on an object (whether or not it floats) is due to the pressure difference between the bottom of the object and the top of the object. If the object is going to be buoyed up, the bottom of the object is subjected to a greater pressure than the top of the object. Remember that pressure is force/area and so, if the pressure at the bottom of an object (submerged under a liquid) is greater than the pressure on top (still in air), there is a net upward force.
This animation shows a block lowered into a liquid and then floating. The density of the block can be changed by click-dragging in the block on the upper left of the animation (click-drag at the white-gray interface of the box). The graph in the lower right of the animation shows the pressure difference (P) as a function of depth (Y). When the net upward force, the buoyant force, is equal to the weight of an object (the white and gray striped block), it floats in the gray liquid. In addition, the amount of fluid displaced when the block is in the fluid is shown in the block to the left.
Assume that the gray liquid is water (density 1000 kg/m3). The density of the block is a fraction of the density of water. So, to start with, the block has a density of 0.4 x 1000 kg/m3 = 400 kg/m3. Note that approximately 40% of the block is submerged when it is floating. Assume the dimensions of the block are 1 m (and it is a cube). First, let's find the net force on the block when it is floating by finding the pressure on the block.
Since pressure as a function of depth in a liquid is ρliquidgy (where ρ is the density of liquid, g is the acceleration due to gravity, and y is the depth in the liquid), what is the pressure at the bottom of the block? It is Patm + ρliquidgy where y = 0.4 m. What is the pressure at the top of the block? It is just Patm. Therefore, the liquid exerts a total force of ΔP A = ρliquidgy A = 400 N, which must also be the block's weight if it is in equilibrium. Why do we neglect the force on the sides of the block due to water pressure?
The animation also shows the spillover of water into a second container (to the left). What is the volume of water in this second container? Using the density of water, what is the weight of the water? Note that this is equal to the buoyant force because, if the block were removed and the water in the spillover tray (to the left) were put back into the main water container, that water would be supported, so the pressure difference supports that weight of water. This means that although the buoyant force is due to the pressure difference at different depths, it is also equal to the weight of the water displaced by the object, as expressed in the equation below:
FB = ρliquidgVliquid displaced = ρliquidgVsubmerged part of object
If the buoyant force is equal to the weight of the object, the object floats. Change the density of the object by click-dragging the mouse on the base of the red arrow in the top left-hand box and then let the animation run. Repeat the calculations to show that the buoyant force is equal to the weight of the object.
What happens if the object is pushed into the water below where it would naturally float (away from equilibrium)? Try using the mouse to drag the floating block farther down in the water. What type of motion do you observe, and why? Think about the forces acting on the block.
Illustration authored by Anne J. Cox.
Applet authored by Fu-Kwun Hwang, National Taiwan Normal University.