Question

You have decided to try to replicate Archimedes task of finding the density of an object by submerging into a liquid and finding the displaced volume. However, you modified the process to be more accurate. First you place the unknown object by placing it on a scale which measures the normal force at 20N. To find the density, you place the object and scale in water and measure the normal force at 17.5N.

(8pts) Draw a force diagram for both situations (one for out of water, one for in the water).

(8 pts) What is the buoyancy force on the object when it is on the scale in the water?

(14 pts) What is the volume and density of the object?

Answer 1

Weight of the body=20 N

Weight of the body in water =17.5 N

Loss of the weight of the body in water= 20 N-17.5N=2.5N

According to Archimedes principle

Loss in weight of the body=weight of the water displaced by the body=2.5N

Relative density of the object =20/2.5=8

Density of the object =relative density of object *density of water

Tension in the string if the object in air=weight of the object =20N

W=mg=20/9.8=0.49kg

Volume of the block V=mass/density=.49/8*1000

=6.125*100000 cube metre

Buoyant force=V*density of water*gravity=6.12500000*1000*9.8=60.025*100000000N