Question

A 18.0 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 5.20 cm below the surface of the water.

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

P₀ = 1.0130  10⁵ N/m².

Enter your answers accurate to at least four significant figures.

F_top= N

F_bottom= N

(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 the top and bottom of the block. (Submit a file with a maximum size of 1 MB.)

Answer 1

(a) Force on the top is due to atmospheric pressure and the pressure due to water column above it.
Area of the surface is 10cm*10cm = 100 cm2 = 0.01m2

h1 = 5.2 cm = 0.052 m

h2 = 5.2+12 = 17.2 cm = 0.172 m

As the force is pressure*area
Ftop = (P0 + h1* *g)* A

here   =density of water = 1000 kg/m3 and A = surface area
Ftop= (1.0130*105 + 0.052*1000*9.8)*0.01

Ftop= 1018.1 N

And on the bottom
Fbottom = (P0 + h2* *g) A
Fbottom= (1.0130*105 + 0.172*1000*9.8)*0.01

Fbottom= 1029.9 N

.......................................................................................

(b)
The scale will read apparent weight, which is (Weight of the block + force on top – force on bottom)

scale reading = 18*9.8 + 1018.1 - 1029.9 = 156.6 N

scale reading = 164.6 N

..................................................................................................
(c)
The buoyant force = Weight of displaces water given by
V**g = (12*10*10)*10-6 * 1000 * 9.8 = 11.8 N

V**g = 11.8 N

Difference = 1029.9 N - 1018.1 N = 11.8 N

Hence proved