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 4.55 cm below the surface of the water.

Figure (a) shows a vertical spring scale. The top of the scale has a hook attached to the ceiling. The bottom of the scale has a hook attached to a rope. The rope hangs down and is attached to a solid block that hangs from the rope. Figure (b) shows the same scale, but now the block is submerged in a tank of water. The reading on the scale in Figure (b) is less than the reading on the scale in Figure (a).

(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

P₀ = 1.0130 10⁵ N/m².

Enter your answers accurate to at least four significant figures.)

F_top= NF_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

Solution: (a) The expression for the pressure at depth is as follows: p= p. + pgh Here, the pressure at a depth is p, pressure at the surface of the liquid is p., density of the liquid is p, depth below the surface of the liquid is h, and the acceleration due to gravity is Substitute 1.0130x10 Nm for p., 1x10 kg/m for p. 9.8m/s for g, and 4.95 cm for h.

P = (1.0130x10 N/mº)+(1x10 kg/m (9.8 m/s -(1.0130x10' N/m)+1x10 kg/m (9.8 m/s )(4.95*10-2 m) =101300+ 485.1 101785 1N/m? The expression for force at the top is, F = P4 -(101785.1Nm?)10cm (10 cm) -10cm 10 cm) = (101785.1N/m) 10x10x10* m ) = 1017.851N The expression for pressure at the bottom is, P: = pi + pgh for P. 1x10 kg/m for p. 9.8 m/s for g, and Substitute 1.0130x10 Nm (4.95+12) em for h. P.-(1.0130-10 Nm*)=(1*10*ks/m')(9.8m/s )|(4.95 +12) cm (10 m) -(1.0130x10' N/m)+(1x10 kg/m (9.8m/s)((4.95+12 )x10- m) =101300+1661.1 102961.1N/m The expression for force at the bottom is, F = P4 - (102961.1Nm*)( 10 an ( 10 cm ( 19 mm =(102961.1N/m?)(10x10x104m) =1029.611N (6) The reading in spring is equal to the apparent weight of the block.

R=mg-F,-F,) = (18.5kg x 9.8 m/s)-(1029.611N-1017.851N) = 169.54N From Archimedes principle buoyant fore is equal to the weight of the water displaced. Fuoyant = Peter 8 =(10)(12cm( " in «10 cm ( in ) 10 cm ( 141 m) (1.5m/s) =11.76N