Question

An 98 kg student who can’t swim sinks to the bottom of the Katutura swimming pool after slipping. His total volume at the time of drowning is 23.3 liters. A rescuer who notices him decides to use a weightless rope to pull him out of the water from the bottom. Use Archimedes’s principle to calculate how much minimum tension (in Newtons) is required in the rope to lift the student without accelerating him in the process of uplift out of the water.  

Answer 1

by using newton's second law in vertical direction,

Fnet = m*a

here, Fnet = net force in vertical direction = T + Fb - Fg

T = tension in rope = ??

Fb = Buoancy force = rho*V*g

Fg = weight of student = m*g

m = mass of student = 98 kg

g = surface gravity = 9.81 m/s^2

rho = density of water = 1000 kg/m^3

V = volume of student = 23.3 ltr = 23.3*10^-3 m^3

a = acceleration in vertical direction = 0 (given)

then, T + 1000*(23.3*10^-3)*9.81 - 98*9.81 = 0

T = 98*9.81 - 23.3*9.81

T = 732.8 N

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