A foam plastic ( sg = 0.58 ) is to be used as a life preserver. What volume of plastic must be used if it is to keep 20 % (by volume) of a 75 kg man above water in a lake? Average density of a man is equal to 1.04 g/cc. Assume the foam plastic’s top surface to be level with the water surface

Let Vp be the required plastic volume.

sgp = plastic specific gravity
sgm = man's specific gravity.
Vm = man's volume = 75000g/1.04 g/cc
= 0.0721 m^3
M = man's mass = 75 kg
sgw = water specific gravity = 1.00

(Total weight) - (buoyancy) = 0
(Vm*sgm + Vp*sgp) - [(Vp*sgw)+0.8*Vm*sqw] = 0

Vp[sgw - sgp] = Vm[sgm - 0.8*sgw]
Vp[0.42] = Vm[1.04 - 0.8]
Vp = 0.57 Vm = 0.0412 m^3

To calculate the volume of foam plastic needed to keep 20% of a 75 kg man above water, we can use the concept of buoyancy.

Buoyancy is the upward force exerted by a fluid on an object immersed in it. It is equal to the weight of the fluid displaced by the object.

Given:
- Specific gravity of the foam plastic (sg) = 0.58
- Weight of the man (W_m) = 75 kg
- Volume fraction required (V_fraction) = 20% = 0.20
- Average density of a man = 1.04 g/cc

Step 1: Convert the man's weight from kg to grams:
W_m = 75 kg × 1000 g/kg = 75,000 g

Step 2: Calculate the weight of the water displaced by the foam plastic:
Weight of water displaced (W_water) = W_m × V_fraction = 75,000 g × 0.20 = 15,000 g

Step 3: Convert the weight of water displaced to volume using the average density of a man:
Volume of water displaced (V_water) = W_water / density = 15,000 g / (1.04 g/cc) = 14,423.08 cc

Step 4: Calculate the volume of the foam plastic needed using the specific gravity:
Volume of plastic needed (V_plastic) = V_water / sg = 14,423.08 cc / 0.58 = 24,888.41 cc

Therefore, approximately 24,888.41 cc or 24.89 liters of foam plastic must be used as a life preserver to keep 20% of a 75 kg man above water in a lake.