Question

An iron anchor weighs 250 pounds in air and has a weight density of 480 lbs/ft3. If it is immersed in sea water that has a weight density of 64 lbs/ft3, how much force would be required to lift it while it is immersed?

Answer 1

Answer:

Explanation:

Given

weight of anchor $W=250\ pounds$

weight density $\rho =480\ lbs/ft^3$

volume of anchor $V=\frac{250}{480}=0.520\ ft^3$

density of sea water $\rho _s=64\ lbs/ft^3$

buoyant force on anchor when immersed in sea water $=\rho _s\times V=33.28 lbs$

Force required to lift the anchor=Weight-Buoyancy

$F=250-33.28=216.72\ lbs$

Answer 2

Answer:

Explanation:

Weight in air = true weight = 250 pounds

density of iron , d = 480 lbs/ft³

density of sea water, d' = 64 lbs/ft³

Volume of iron = mass / density of iron

V = 250 / 480 = 0.521 ft³

Force required to lift = weight of iron in water = Apparent weight of iron

Weight in water = true weight - buoyant force

Weight in water = 250 x g - volume x density of water x g

Weight in water = 250 g - 0.521 x 64 x g

Weight in water = 216.67 g

Weight in water = 216.67 pounds

Thus, the force required to lift the iron is 216.67 pounds.