Question

A 18.5 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 4.95 cm below the surface of the water. (a) What are the magnitudes of the forces (in N) acting on the top and on the bottom of the block due to the surrounding water? (Use P. - 1.0130 X 10' N/m². Enter your answers accurate to at least four significant figures.) Foop C N Fbottom (b) What is the reading of the spring scale (in N)? N (c) Show that the buoyant force equals the difference between the forces at th top and bottom of the block. (Submit a file with a maximum size of 1 MB.) Choose Fto no file selected This answer has not been graded yet.

Answer 1

Solution (9) A= Area altop and bottem face = 0.01 m2 Ftop= Po At hp pga = 110130x10f Xocol + 4095X10 Exlooo x 9.8 x = 1013.0 + 4.851 = 1017851 N F botten = (Pot hz Pg) A . where V=Volume = 0.0012 m Fbottom= [110/308/04+ 1609531631000x91874 =T102961.1 X 0.01] N = 1029. 61.1 N Ftop 10 17.851 IN F boltem = 1 To 29,611 7N s reading = 18,5X968 +1017.851 – 1029611 169.54 H FB = veg = 0.0012X1000X918=11176 Fbottom – Hop= 11:76 .So verified O = 169.54 N ©