A 20.0 kg block of metal measuring 12.0 cm by 10.0 cm by 10.0 cm is suspended fr

Question

A 20.0 kg block of metal measuring 12.0 cm by 10.0 cm by 10.0 cm is suspended from a scale and immersed in water as shown in the figure below. The 12.0 cm dimension is vertical, and the top of the block is 4.70 cm below the surface of the water. (a) What are the magnitudes of the forces acting on the top and on the bottom of the block due to the surrounding water? (Use P_0 = 1.0130 times 10^5 N/m^2.) (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block.

Explanation / Answer

a] Ftop = Pressure*area = [Po+ rho g h]*area = [101300+ 1000*9.8*0.047]*0.10*0.10 = 1017.6 N

Fbottom =  [Po+ rho g h]*area = [101300+ 1000*9.8*(0.047+0.12)]*0.10*0.10 = 1029.4 N

b] Reading in the spring scale = 20*9.8 + 1017.6-1029.4 = 184.2 N answer

c] Buyant Force + Spring force = mg

Buyant Force = mg - Spring force

= mg - [ mg + Ftop - Fbottom]

= Fbottom - Ftop

Hence shown

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