A diving bell has a weight of 30000 N when measured on shore. In the water, suspended by a cable, the tension in the cable is only 12000 N. What is the buoyant force, FB, on the diving bell when it is in the water?
30000-12000=18000
To find the buoyant force on the diving bell when it is in water, we can use Archimedes' principle, which states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
First, let's calculate the apparent weight of the diving bell in water. The apparent weight is the difference between the weight of the diving bell on land and the tension in the cable.
Apparent weight in water = Weight on land - Tension in the cable
Weight on land = 30000 N (given)
Tension in the cable = 12000 N (given)
Apparent weight in water = 30000 N - 12000 N = 18000 N
The apparent weight in water represents two forces acting on the diving bell: its actual weight and the buoyant force. Since the diving bell is in equilibrium, the magnitude of the buoyant force is equal to the magnitude of the apparent weight.
Therefore, the buoyant force, FB, on the diving bell when it is in the water is 18000 N.