A 8.10 kg rock whose density is 4100 kg/m3 is suspended by a string such that half of the rock's volume is under water.
What is the tension in the string?
Given,
m = 8.1 kg ; rho = 4100 kg/m^3
V = m/d = 8.1/4100 = 1.98 x 10^-3 m^3
the forces on rock are:
W = mg
W = 8.1 x 9.8 = 79.38 N
Fb = V rho g
Volume of dispalced water:
V' = V/2 = 1.98 x 10^-3/2 = 9.9 x 10^-4 m^3
Fb = 9.9 x 10^-4 x 1000 x 9.8 = 9.68 N
Tenstion T which acts upwards,under equilibrium
Fnet = 0
Fb + T - W = 0
T = W - Fb
T = 79.38 - 9.68 = 69.7 N
Hence, T = 69.7 N
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