Answer to Question #173222 in Physics for Isaac Addai

Question

A piece of metal, mass 3.6 kg is suspended from a spring balance. What is the reading of the balance (in N) a.      With the metal in air [36N]

b.     With the metal in water [32N]

c.      With the metal in brine [31.2N]

d.     What is the density of a liquid in which it gives a reading of 33N? [750kg/m_­­­­­³].

(Density of brine 1200 kg/m³; density of metal=900kg/m³)

Answer 1

(a)

$$N=mg=3.6\ kg\cdot10\ \dfrac{m}{s^2}=36\ N.$$

(b)

$$N=mg-F_B=\rho_mV_mg-\rho_wV_wg,$$ $$N=(\rho_mV_m-\rho_wV_w)g=\rho_mV_mg(1-\dfrac{\rho_w}{\rho_m}),$$ $$N=mg(1-\dfrac{\rho_w}{\rho_m}),$$ $$N=3.6\ kg\cdot10\ \dfrac{m}{s^2}\cdot(1-\dfrac{1000\ \dfrac{kg}{m^3}}{9000\ \dfrac{kg}{m^3}})=32\ N.$$

(c)

$$N=mg(1-\dfrac{\rho_w}{\rho_b}),$$ $$N=3.6\ kg\cdot10\ \dfrac{m}{s^2}\cdot(1-\dfrac{1200\ \dfrac{kg}{m^3}}{9000\ \dfrac{kg}{m^3}})=31.2\ N.$$

(d)

$$N=mg(1-\dfrac{\rho_L}{\rho_m}),$$ $$\rho_L=(1-\dfrac{N}{mg})\rho_m,$$ $$\rho_L=(1-\dfrac{33\ N}{3.6\ kg\cdot10\ \dfrac{m}{s^2}})\cdot9000\ \dfrac{kg}{m^3}=750\ \dfrac{kg}{m^3}.$$