Answer to Question #88016 in Mechanics | Relativity for gc

Question

A scale reads 274 N when a piece of brass is hanging from it. What does it read (in N) when it is lowered so that the brass is submerged in water?

Answer 1

Let's first find the mass of the piece of brass from the formula:

$$W = mg,$$

here, $W = 274 N$ is the weight of the piece of brass in air, $m$ is the mass of the piece of brass and $g = 9.8 m/s^2$ is the acceleration due to gravity.

Then, we get:

$$m = \dfrac{W}{g} = \dfrac{274 N}{9.8 \dfrac{m}{s^2}} = 27.9 kg.$$

Let's also find the volume of the piece of brass from the definition of the density:

$$\rho = \dfrac{m}{V},$$

here, $\rho = 8553 kg/m3$ is the density of the brass, $V$ is the volume of the piece of brass.

Then, we get:

$$V = \dfrac{m}{\rho} = \dfrac{27.9 kg}{8553 \dfrac{kg}{m^3}} = 3.26 \cdot 10^{-3} m^3.$$

Let’s assume the upwards to be positive direction and the apparent weight of the piece of brass in water directed downward, the force of gravity that acts on the piece of brass (or weight) directed downward and the buoyant force directed upward:

$$-W_{App} = F_{B} - W,$$ $$W_{App} = W - F_{B} ,$$

here, $W_{App}$ is the apparent weight of the piece of brass submerged in water, $F_{B} = \rho_{w}V_{w}g = \rho_{w}Vg$ is the buoyant force acting on the piece of brass, $\rho_{w} = 1000 kg/m^3$ is the density of the water and $V_{w}$ is the volume of displaced water by the piece of brass.

Finally, from this formula we can find the read of the scale when peace of brass submerged in water (or its apparent weight):

$$W_{App} = W - \rho_{w}Vg,$$ $$W_{App} = 274 N - 1000 \dfrac{kg}{m^3} \cdot 3.26 \cdot 10^{-3} m^3 \cdot 9.8 \dfrac{m}{s^2} = 242 N.$$

Answer:

$W_{App} = 242 N.$