An inverted cone is placed in a water tank as shown (above left). if the weight of the cone is 16.5 n, what is the tensile force in the cord connecting the cone to the bottom of the tank?

T = 365246.01 N   (↑)

Explanation:

We must show the image, in order to explain the solution.

Then, we apply

∑ Fy = 0 (+↑)

Ff - W - T = 0  ⇒  T = Ff - W

We can get the the buoyant force (Ff) as follows:

Ff = ρ*g*V

where V is obtained using the equation

V = (1/3)*π*r²*h

we get r as follows

(20 cm / 30 cm) = (r / 20 cm)   ⇒  r = 13.33 cm = 0.133 m

h = 20 cm = 0.20 m

then

V = (1/3)*π*r²*h = V = (1/3)*π*(0.133 m)²*(0.20 m) = 37.23 m³

then we get the buoyant force (Ff)

Ff = ρ*g*V = (1000 Kg/m³)*(9.81 m/s²)*(37.23 m³) = 365262.51 N

Finally, we use the equation

T = Ff - W = 365262.51 N - 16.5 N = 365246.01 N   (↑)

the answer would be a

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