Question

2 kg bras black is attached to a string and submerged under water. Find the buoyant force and the tension in the rope. ( density of bras=8700 kg/m^3

Answer 1

Question#18738

2 kg bras black is attached to a string and submerged under water. Find the buoyant force and the tension in the rope. (density of bras=8700 kg/m^3)

Solution:

According to Archimedes' principle:

$F_{buoyant} = \rho gV$, were $\rho$ - density of water, $V$ - value of an object, $g = 9.8 \, \text{m/s}^2$

Such as: $m = \rho_{bras} V$, $V = \frac{m}{\rho_{bras}}$

$$F_{buoyant} = \frac{\rho g m}{\rho_{bras}} = \frac{1000 \times 9.8 \times 2}{8700} = 2.25 \, \text{N}$$

The tension in the rope is:

$$F = m g - F_{buoyant}$$ $$F = 2 \times 9.8 - 2.25 = 17.35 \, \text{N}$$

Answer: the buoyant force- 2.25 N, the tension on the rope- 17.35 N.