Liquid exerts a force on objects immersed or floating in it. This force is equal to the weight of the liquid that is displaced by an object. This is also known as Archimedes’ principle. The unit for the buoyant force (like other forces) is the Newton (N).

*buoyant force =(density of liquid)(gravitational acceleration)(volume of liquid)*

*= (density)(gravitational acceleration)(height of liquid)(surface area of object)*

*F _{b}* = ρgV = ρghA

*F _{b}* = buoyant force of a liquid acting on an object (N)

*ρ* = density of the liquid(kg/m^{3})

*g* = gravitational acceleration(9.80 m/s^{2})

*V* = volume of liquid displaced (m^{3} or liters, where 1 m^{3} = 1000 L)

*h* = height of water displaced by a floating object(m)

*A* = surface area of a floating object(m^{2})

**Buoyancy Formula Questions:**

1) A golden crown has been placed in a tub of water. The volume of water displaced is measured to be 1.50 liters. The density of water is 1000 kg/m^{3}, or 1.000 kg/L. What is the buoyant force acting on the crown?

**Answer:** The buoyant force can be found using the formula. First, we ensure that the units used for volume are the same. If 1 m^{3} = 1000 L, then 1.50 L = 0.00150 m^{3}. The buoyant force is:

*F _{b} = ρgV*

*F _{b}* = (1000 kg/m

^{3})(9.80 m/s

^{2})(0.00150 m

^{3})

*F _{b}* = 14.7 kg∙m/s

^{2}

*F _{b}* = 14.7 N

The buoyant force acting on the golden crown is 14.7 N.

2) An empty canoe is floating by a dock in a lake. The bottom of the canoe has a surface area of 2.70 m^{2}. With nothing in it, the canoe displaces only 2.00 cm (0.0020 m) of water. Then, the canoe is filled up with gear for a camping trip. Once the canoe is fully loaded, it displaces 30.0 cm (0.300 m) of water. How much more buoyant force is exerted on the canoe by the water when it is fully loaded, versus when it is empty?

**Answer:** The difference between the buoyant forces can be found by solving for the buoyant forces before and after loading the canoe. The density of water is 1000 kg/m^{3}, or 1.000 kg/L. First, the buoyant force for the empty canoe:

*F _{b,empty} = ρgh_{empty} A*

*F _{b,empty}* = (1000 kg/m

^{3})(9.80 m/s

^{2})(0.0020 m)(2.70 m

^{2})

*F _{b,empty}* = 52.92 kg∙m/s

^{2}

*F _{b,empty}* = 52.92 N

Now, the buoyant force for the loaded canoe:

*F _{b,loaded} = ρgh_{loaded} A*

*F _{b,loaded}* = (1000 kg/m

^{3})(9.80 m/s

^{2})(0.300 m)(2.70 m

^{2})

*F _{b,loaded}* = 7938 kg∙m/s

^{2}

*F _{b,loaded}* = 7938 N

Now, find the difference between these values (the symbol “∆” means “the change in”):

*∆F = F _{b,loaded} – F_{b,empty}*

*∆F* = 7938 N – 52.92 N

*∆F* = 7885.08 N

*∆F* ≈ 7885 N

Rounded to four significant figures, the difference in buoyant force between the empty canoe and when it is loaded is 7885 N.