Question

The weight of a rectangular block of low-density material is$\mathbf{15}\mathbf{\text{\hspace{0.17em}N}}$. With a thin string, the center of the horizontal bottom face of the block is tied to the bottom of a beaker partly filled with water. When$\mathbf{25}\mathbf{\%}$of the block’s volume is submerged, the tension in the string is$\mathbf{10}\mathbf{\text{\hspace{0.17em}N}}$.

(a) Find the buoyant force on the block.

(b) Oil of density$\mathbf{800}\mathbf{\text{\hspace{0.17em}kg}}\mathbf{/}{\mathbf{\text{m}}}^{\mathbf{\text{3}}}$is now steadily added to the beaker, forming a layer above the water and surrounding the block. The oil exerts forces on each of the four sidewalls of the block that the oil touches. What are the directions of these forces?

(c) What happens to the string tension as the oil is added? Explain how the oil has this effect on the string tension.

(d) The string breaks when its tension reaches$\mathbf{60}\mathbf{\text{\hspace{0.17em}N}}$. At this moment,$\mathbf{25}\mathbf{\%}$of the block’s volume is still below the water line. What additional fraction of the block’s volume is below the top surface of the oil?

Short answer

(a) The buoyant force on the block by water is $\text{25\hspace{0.17em}N}$.

(b) The force from the oil on the sidewalls of the block are directed into the block.

(c) As oil is added the string tension increases. This is because of the increase in buoyancy due to increase in liquid level.

(d) An additional $62.5\%$of the volume of the block is below the top surface of the oil.

Step-by-step solution

Identification of the given data:

The given data can be listed below.

• Weight of the block, $W=15\text{\hspace{0.17em}N}$
• Tension in the string in the presence of just water, $T_w=10\text{\hspace{0.17em}N}$
• Percentage of volume of block submerged in water is $25\%$.
• Tension at which string breaks, $T_b=60\text{\hspace{0.17em}N}$
• Density of oil, $d_o=800\raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.$

Significance of the Buoyancy:

The buoyant force from a liquid of density $\mathit{d}$when volume of the liquid displaced$\mathit{v}$ is

$\mathit{B}\mathbf{=}\mathit{d}\mathit{g}\mathit{v}$ ...(1)

Here, $\mathit{g}$is the acceleration due to gravity having value,

$\mathit{g}\mathbf{=}\mathbf{9}\mathbf{.8}\raisebox{1ex}{${\displaystyle \mathbf{\text{m}}}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{\text{s}}}^{\mathbf{\text{2}}}$}\right.$

(a) Determining the buoyancy force on the block in the absence of oil:

Since the block is in equilibrium the net force on it is zero. The only force upward is the buoyancy. The forces downward are the weight of the block and the tension in the string. Thus the buoyancy $B$is,

$B=W+T_w$

Substitute all the value in the above equation,

$\begin{array}{rcl}B& =& 15\text{\hspace{0.17em}N}+10\text{\hspace{0.17em}N}\\ & =& 25\text{\hspace{0.17em}N}\end{array}$

Hence, the buoyancy is $25\text{\hspace{0.17em}N}$.

(b) Determining the direction of forces from the oil:

The forces from the oil on the four sidewalls of the block are directed into block. These forces cancel each other.

(c) Determining the change in string tension upon addition of oil:

Buoyancy is caused due to the difference in pressure. The oil by itself doesn't apply any force directed upward on the block. But the oil increases the depth of the liquid medium surrounding the block. The pressure difference increases which in turn increases the buoyancy on the block. To balance it the string tension also increases.

(d) Determining the fractional volume of block submerged in oil when the string breaks:

Let the total volume of the block be $V$. $25\%$of this volume is underwater. Thus $0.25\text{V}$is the volume of displaced water.

Density of water, $d_w=1000\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$.

From equation (1),

$$\begin{gathered} B=d_{w}g\times 0.25V \\ V=\frac{B}{0.25d_{w}g} \end{gathered}$$

Substitute all the value in the above equation,

$\begin{array}{rcl}V& =& \frac{25\text{\hspace{0.17em}N}}{0.25\times 1000{\displaystyle \raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.}\times 9.8{\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.}}\\ & =& 0.0102\cdot \left(\frac{1{\displaystyle \raisebox{1ex}{$\text{kg}\cdot \text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.}}{1\text{\hspace{0.17em}kg}}\right)\cdot \left(\frac{1{\text{\hspace{0.17em}m}}^{\text{3}}}{1{\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.}}\right)\\ & =& 0.0102{\text{\hspace{0.17em}m}}^{\text{3}}\end{array}$

Let the fraction of volume of block submerged in oil be $f$. Thus, $fV$is the volume of displaced oil.

The new force equation is,

$$\begin{gathered} B+d_{o}gfV=W+T_b \\ f=\frac{W+T_b-B}{d_{o}gV} \end{gathered}$$

Substitute all the value in the above equation,

role="math" localid="1663756446920" $\begin{array}{rcl}f& =& \frac{15\text{\hspace{0.17em}N}+60\text{\hspace{0.17em}N}-25\text{\hspace{0.17em}N}}{800{\displaystyle \raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.}\times 9.8{\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.}\times 0.0102{\text{\hspace{0.17em}m}}^{\text{3}}}\\ & =& 0.625\cdot \left(1\text{\hspace{0.17em}N}\times \frac{1{\displaystyle \raisebox{1ex}{$\text{kg}\cdot \text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.}}{1\text{\hspace{0.17em}N}}\right)\cdot \left(\frac{1}{1{\displaystyle \raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.}}\right)\cdot \left(\frac{1}{1{\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.}}\right)\cdot \left(\frac{1}{1{\text{\hspace{0.17em}m}}^{\text{3}}}\right)\\ & =& 0.625\\ & =& 62.5\%\end{array}$

Hence, the required percentage is $62.5\%$.