Question

Submerged Sphere in a Beaker

A cylindrical beaker of height 0.100m and negligible weight is filled to the brim with a fluid of density

? = 890kg/m3 . When the beaker is placed on a scale, its weight is measured to be 1.00N .(Figure 1)

A ball of density ?b = 5000kg/m3 and volume V = 60.0cm3 is then submerged in the fluid, so that some of the fluid spills over the side of the beaker. The ball is held in place by a stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration due to gravity is g = 9.81m/s2 .

1. What is the weight Wb of the ball?

Express your answer numerically in newtons.

2. What is the reading W2 of the scale when the ball is held in this submerged position? Assume that none of the water that spills over stays on the scale.

Calculate your answer from the quantities given in the problem and express it numerically in newtons.

3.

What is the force Fr applied to the ball by the rod? Take upward forces to be positive (e.g., if the force on the ball is downward, your answer should be negative).
Express your answer numerically in newtons.

4. The rod is now shortened and attached to the bottom of the beaker. The beaker is again filled with fluid, the ball is submerged and attached to the rod, and the beaker with fluid and submerged ball is placed on the scale.
What weight W3 does the scale now show?

Answer 1

Concepts and reason

The concepts required to solve the given problem are Archimedes Principle and weight.

Calculate the weight of the ball by taking the product of mass and acceleration due to gravity.

Calculate the reading on scale with the help of equilibrium condition of force.

Fundamentals

Weight is the force with which the earth attracts every object towards itself.

The weight of an object is given by,

$W = mg$

Here, $m$ is the mass of object and $g$ is the acceleration due to gravity.

Archimedes Principle states that the upward force exerted on the body immersed in a fluid is equal to the weight of the fluid that is displaced.

The buoyant force can be calculated by the formula,

${F_{{\rm{buoyancy}}}} = \rho Vg$

Here, $\rho$ is the density of water, $V$ is the volume of water displaced by the ball and $g$ is the acceleration due to gravity.

The equilibrium condition for force gives,

$\Sigma \vec F = 0$

Here, $F$ is force

(1)

The weight of the ball can be found by the formula,

${W_{{\rm{ball}}}} = {m_{{\rm{ball}}}}g$ …… (1)

Here, ${m_{{\rm{ball}}}}$ is the mass of ball and $g$ is the acceleration due to gravity.

The density, mass and volume of an object are related as,

$\rho = \frac{{{m_{{\rm{ball}}}}}}{V}$

Here, $\rho$ is the density, ${m_{{\rm{ball}}}}$ is the mass of ball and $V$ is the volume.

Rearrange the above equation.

${m_{{\rm{ball}}}} = \rho V$

Substitute $5000{\rm{ kg / }}{{\rm{m}}^3}$ for $\rho$and $60.0{\rm{ c}}{{\rm{m}}^3}$ for $V$ in the above equation.

$\begin{array}{c}\\{m_{{\rm{ball}}}} = \left( {5000{\rm{ kg / }}{{\rm{m}}^3}} \right)\left( {60{\rm{ c}}{{\rm{m}}^3}} \right)\left( {\frac{{1{\rm{ }}{{\rm{m}}^3}}}{{{{10}^6}{\rm{ c}}{{\rm{m}}^3}}}} \right)\\\\ = 0.3{\rm{ kg}}\\\end{array}$

Substitute $0.3{\rm{ kg}}$ for ${m_{{\rm{ball}}}}$ and $9.81{\rm{ m / }}{{\rm{s}}^2}$ for $g$ in equation (1).

$\begin{array}{c}\\{W_{{\rm{ball}}}} = \left( {0.3{\rm{ kg}}} \right)\left( {9.81{\rm{ m / }}{{\rm{s}}^2}} \right)\\\\ = 2.94{\rm{ N}}\\\end{array}$

(2)

According to Archimedes’ principle, the weight of the beaker as shown by the scale will be,

${W_{{\rm{beaker}}}} = W\left( {{\rm{water}} - {\rm{spilled water}}} \right) + W\left( {{\rm{object}}} \right)$ …… (2)

But, $W\left( {{\rm{object}}} \right) = W\left( {{\rm{spilled water}}} \right)$

Substitute $W\left( {{\rm{spilled water}}} \right)$ for $W\left( {{\rm{object}}} \right)$ in equation (2).

$\begin{array}{c}\\{W_{{\rm{beaker}}}} = W\left( {{\rm{water}} - {\rm{spilled water}}} \right) + W\left( {{\rm{spilled water}}} \right)\\\\ = W\left( {{\rm{water}}} \right)\\\end{array}$

When the ball is held in this submerged position the reading on the scale will be equal to the weight of the beaker with water.

Substitute $1.0{\rm{ N}}$ for $W\left( {{\rm{water}}} \right)$ in the above equation.

${W_{{\rm{beaker}}}} = 1.0{\rm{ N}}$

(3)

The force applied to the ball by the rod can be found by calculating the net force on the ball.

$\sum {{F_{{\rm{net}}}}} = {W_{{\rm{ball}}}} - {F_{{\rm{buoyancy}}}} - {F_{{\rm{rod}}}}$

Here, ${W_{{\rm{ball}}}}$ is the weight of the ball, ${F_{{\rm{buoyancy}}}}$ is the buoyant force and ${F_{{\rm{rod}}}}$ is the force on the ball by the rod.

Since the ball is stationary, the net force on the ball will be zero.

$0 = {W_{{\rm{ball}}}} - {F_{{\rm{buoyancy}}}} - {F_{{\rm{rod}}}}$

${F_{{\rm{rod}}}} = {W_{{\rm{ball}}}} - {F_{{\rm{buoyancy}}}}$ …… (3)

The buoyant force can be calculated by the formula,

${F_{{\rm{buoyancy}}}} = \rho Vg$

Here, $\rho$ is the density of water, $V$ is the volume of water displaced by the ball and $g$ is the acceleration due to gravity.

Substitute $890{\rm{ kg / }}{{\rm{m}}^3}$ for $\rho$, $9.81{\rm{ m / }}{{\rm{s}}^2}$ for $g$ and $0.60{\rm{ c}}{{\rm{m}}^3}$ for $V$ in the above equation.

$\begin{array}{c}\\{F_{{\rm{buoyancy}}}} = \left( {890{\rm{ kg / }}{{\rm{m}}^3}} \right)\left( {60{\rm{ c}}{{\rm{m}}^3}} \right)\left( {\frac{{1{\rm{ }}{{\rm{m}}^3}}}{{{{10}^6}{\rm{ c}}{{\rm{m}}^3}}}} \right)\left( {9.81{\rm{ m / }}{{\rm{s}}^2}} \right)\\\\ = 0.524{\rm{ N}}\\\end{array}$

Substitute $2.94{\rm{ N}}$ for ${W_{ball}}$ and $0.524{\rm{ N}}$ for ${F_{buoyancy}}$ in equation (3).

$\begin{array}{c}\\{F_{{\rm{rod}}}} = \left( {2.94{\rm{ N}}} \right) - \left( {0.524{\rm{ N}}} \right)\\\\ = 2.42{\rm{ N}}\\\end{array}$

(4)

When the rod is attached to the bottom of the beaker the weight of the beaker as shown by the scale will be,

$W' = {W_{{\rm{water}}}} + {F_{{\rm{rod}}}}$

Here, ${W_{{\rm{water}}}}$ is the weight of water and ${F_{{\rm{rod}}}}$ is the force applied by the rod on the ball.

Substitute $1{\rm{ N}}$ for ${W_{{\rm{water}}}}$ and $2.42{\rm{ N}}$ for ${F_{{\rm{rod}}}}$ in the above equation.

$\begin{array}{c}\\W' = 1.00{\rm{ N}} + 2.42{\rm{ N}}\\\\ = {\rm{3}}{\rm{.42 N}}\\\end{array}$

Ans: Part 1

The weight of the ball was found to be $2.94{\rm{ N}}$.

Part 2

The reading on the scale when the ball is held in the submerged position is $1.0{\rm{ N}}$.

Part 3

The force applied to the ball by the rod is $2.42{\rm{ N}}$.

Part 4

The weight shown on the scale when the rod is shortened and attached to the bottom of the beaker is $3.42{\rm{ N}}$.