A 10.5 kg block of metal is suspended from a scale and immersed in water, as in

Question

A 10.5 kg block of metal is suspended from a scale and immersed in water, as in the figure below. The dimensions of the block are 12.0 cm × 10.0 cm × 10.0 cm. The 12.0 cm dimension is vertical, and the top of the block is 5.00 cm below the surface of the water. Mg (a) what are the forces exerted by the water on the top and bottom of the block? Take Po = 1.0130 x 105 N/m2 Ftop bottom (b) What is the reading of the spring scale? Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. N (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block. (Do this on paper. Your instructor may ask you to turn in this work.)

Explanation / Answer

Given,

m = 10.5 kg ; 912,10,10 ; d = 5 cm

a)P0 = 1.013 x 10^5 N/m^2

P(top) = P0 + P

P = rho g h

P(top) = P0 + rho g h

P(top) = 1.013 x 10^5 + 1000 x 9.81 x 5 x 10^-2 = 101790.5 Pa

P(top) = 101790.5 Pa = 1.0179 x 10^5 Pa

P = F/A => F = PA

F(top) = 1.0179 x 10^5 Pa x 0.1 = 1017.9 N

Hence, F(top) = 1017.9 N

P(bottom) = P0 + rho g H

P(bottom) = 1.013 x 10^5 + 1000 x 9.81 x 17 x 10^-2 = 102967.7 Pa

P(bottom) = 102967.7 Pa = 1.02967 x 10^5 Pa

F(bottom) = 1.02967 x 10^5 x 0.1 = 1029.7 N

Hence, F(bottom) = 1029.7 N

b)The net force in Y direction is:

Fnet = T + F(bottom) - F(top) - mg = 0

T = mg - [F(bottom) - F(top)]

T = 10.5 x 9.81 - [1029.7 - 1017.9] = 91.21 N

Hence, T = 91.21 N

c)We know that

Fb = rho V g

Fb = 1000 x (0.1^2 x 0.12) x 9.81 = 11.8 N

[F(bottom) - F(top)] = [1029.7 - 1017.9] = 11.8 N

Hence, Proved.

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