Answer to Question #287369 in Physics for Aida

Question

A 10 kg block of metal measuring 12 cm x 10 cm x 10 cm is suspended

from a scale and immersed in water. The 12 cm dimension is vertical, and

the top of the block is 5 cm below the surface of the water as shown in the

figure.

(a) What are the forces acting on the top and on the bottom of the block?

(b) What is the reading of the spring scale? 86.2 N

(c) Show that the buoyant force equals the difference between the forces

at the top and bottom of the block.

Answer 1

(a) The forces acting on the top and bottom of the block:

$$F_t=p_\text{water, top}·A=\\=\rho gha^2=1000·9.8·0.05·0.1^2=4.9\text{ N}.\\\space\\ F_b=p_\text{water, bottom}·A=\\=(\rho g(h+y))a^2=\\=[1000·9.8·(0.12+0.05)]·0.1^2=\\=16.66\text{ N}.\\$$

(b) The reading of the spring scale:

$$R=W-(F_b-F_t)=mg-(F_b-F_t)=86.24\text{ N}.$$

(c) If we find the value $F_b-F_t$, we will notice it precisely equals a value of $\rho_\text{water}gV_\text{block}.$