An anchor made of iron weighs 833 N on the deck of a ship. If the anchor is now suspended in seawater by a massless chain, what is the tension in the chain? (Note: The density of iron is 7800 kg/m3 and the density of seawater is 1025 km/m3)

723.54 N

Explanation:

Weight of anchor = 833 N

density of iron = 7800 mg/m^3

density of sea water = 1025 kg/m^3

According to the Archimedes principle, if a body is immersed wholly or partly in a liquid it experiences a loss in weight of the body which is equal to the weight of liquid displaced by the body.

The loss in the weight of body is equal to the buoyant force acting on the body.

The formula for the buoyant force acting on the body

= volume of body x density of liquid x acceleration due to gravity

Weight of anchor = mass x acceleration due to gravity

833 = m x 9.8

m = 85 kg

mass of anchor = Volume of anchor x density of iron

85 = V x 7800

V = 0.01089 m^3

Buoyant force on the anchor = Volume of anchor x density of sea water x g

= 0.01089 x 1025 x 9.8 = 109.46 N

So, the tension in the chain = Apparent weight of the anchor

= Weight - Buoyant force

= 833 - 109.46 = 723.54 N

Thus, the tension in the chain is 723.54 N.