A 14.5 kg block of metal measuring 12.0 cm by 10.0 cm by 10.0 cm is suspended fr

Question

A 14.5 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 5.25 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. (Do this on paper. Your instructor may ask you to turn in this work.)

Explanation / Answer

Ptop = Po+rho_water *g * h_top

h_top = 5.25 cm = 5.25 x 10^-2 m

Ptop = 1.013 x 10^5 + ( 1000 * 9.8 * 5.25 x 10^-2)

Ptop = 101814.5 Pa

Pbottom = Po + rho_water * g* h_bottom

h_bottom = 12+5.25 = 17.25 cm = 17.25 x 10^-2 m

Pbottom = 102990.5 Pa

we know F = P *A

A = area = 10 cm x 10cm = 0.01 m^2

Ftop = Ptop*A = 1018.145 N

Fbottom = Pbottom*A = 1029.905 N

part b )

block is in equilibrium

sum of all forces in y direction = 0

Fy = T+Fbot - Ftop - mg = 0

T = 130.34 N ( equal to scale reading )

part c )

From Archimedes’s principle, the buoyant force on the block equals the weight of the displaced water

B = rho*water * V * g

V = volume of block = 12 x 10^2 * 10 x 10^2 * 10x 10^-2 = 1.2 x 10^-3

B = 11.76 N

Fnet = Fbotttom - Ftop = 11.76 N

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