Question

Suppose that you suspend a 5 kg object on a spring scale and submerge the whole object in a liquid. The apparent weight on the spring scale reads 10 N. What is the buoyant force acting on the object? (10 points)

Answer 1

39 N is the buoyant force acting on the object.

• Archimedes' principle is used to solve our problem. It states that, “The upward buoyant force that is exerted on a body immersed in a fluid, whether partially or fully submerged, is equal to the weight of the fluid that the body displaces and acts in the upward direction at the center of mass of the displaced fluid”.

• Apparent weight = Weight of the object - Buoyant force

In the given problem, Apparent weight = 10 N

• 10 N = Weight of the object - Buoyant force

The weight of the object can be calculated using the formula:

• Weight = mass × acceleration due to gravity

Assume the acceleration due to gravity is 9.8 m/s². So,

• Weight of the object = 5 kg × 9.8 m/s²

• Weight of the object = 49 N

Substituting, we have:

• 10 N = 49 N - Buoyant force

• Hence, Buoyant force = 39 N

Answer 2

39 N or 40 N

Step-by-step explanation:

considering average value of acceleration due to gravity (g)=9.8m/sq. second

wt of 5 kg object = mg

= 5*9.8

= 49N

so, upthrust(buoyant force) = 49-10 = 39N

if you consider g= 10m/sq. sec

then wt of object = 50 N and upthrust= 40 N