# A tank containing oil of (sp gr=0.8) rests on a scale and weighs 78.6 N .By mean of a wire,a 6.0 cm cube of aluminum, (sp gr=2.7) is submerged in the oil? . Find a) the tension in the wire b)the scale reading none of the oil overflows Ans; (a) 4N (b)80N

Apr 21, 2018

a) ${F}_{t} = 4.026 N$
b) ${F}_{w} = 80.295 N$
See explanation

#### Explanation:

$\text{PART A}$

First let's find the buoyant force on the cube of aluminum. To do this, we can use Archimedes' principle:

${F}_{b} = {W}_{f}$

Where ${F}_{b}$ is the buoyant force and ${W}_{f}$ is the weight of the displaced fluid.

Unfortunately, we do not know the weight of the displaced fluid, so we must use another form of his principle.

${F}_{b} = \rho {V}_{f} g$

Where $\rho$ is the density of the fluid, ${V}_{f}$ is the volume of the displaced fluid, and $g$ is acceleration due to gravity.

Since the cube is submerged in the oil, the volume of the cube is equal to the volume of the displaced fluid.

${V}_{f} = {6}^{3} \cancel{c {m}^{3}} \cdot \frac{1 {m}^{3}}{1000000 \cancel{c {m}^{3}}} = 0.000216 {m}^{3}$

Using specific gravity to find the density of oil:

$\text{S.G.} = \frac{{\rho}_{o i l}}{{\rho}_{w a t e r}}$

.8=(rho_(oil))/(1000(kg)/m^3

${\rho}_{o i l} = 800 \frac{k g}{m} ^ 3$

Therefore:

${F}_{b} = 800 \frac{k g}{\cancel{{m}^{3}}} \cdot 0.000216 \cancel{{m}^{3}} \cdot 9.81 \frac{m}{s} ^ 2$

${F}_{b} = 1.695 N$

Now that we have the force applied upwards by the oil onto the aluminum cube, we need to find the force exerted by the cube downwards.

We need to find the mass of the cube using its specific gravity:

"S.G."=(rho_("aluminum"))/(rho_(water))

2.7=(rho_("aluminum"))/(1000(kg)/m^3

${\rho}_{\text{aluminum}} = 2700 \frac{k g}{m} ^ 3$

Thus the weight of the block of aluminum is

${F}_{w} = {\rho}_{\text{aluminum}} {V}_{c u b e} g$

${F}_{w} = 2700 \frac{k g}{\cancel{{m}^{3}}} \cdot 0.000216 \cancel{{m}^{3}} \cdot 9.81 \frac{m}{s} ^ 2$

${F}_{w} = 5.721 N$

The net force on the cube is

${F}_{\text{net}} = {F}_{b} - {F}_{w}$

${F}_{\text{net}} = 1.695 N - 5.721 N = - 4.026 N$

The tension on the wire is the force needed to keep the block stationary, so it is

${F}_{t} = 4.026 N$

$\text{PART B}$

It is given that the weight of the tank and the oil is 78.6 N. From here, realize that since the cube is not weighing the solution down because the wire is not attached to the tank. You cannot simply add the weight of the cube to the measurement in this case.

What you have to do is think back to Newton's Third Law: "every action has an equal and opposite reaction" If the fluid is pushing up against the cube by the buoyant force, the fluid must also be pushing down on the bottom of the tank with an equal and opposite reaction force in order to maintain equilibrium.

To find the final reading on the scale, simply add the buoyant force of the oil to the original weight:

${F}_{w f} = {F}_{w i} + {F}_{b}$

${F}_{w f} = 78.6 N + 1.695 N = 80.295 N$

The final reading on the scale should be 80.295 N.