Chapter 8, Problem 8.82CP

(a)

To determine

The expression for $H$ as a function of $F$.

Answer to Problem 8.82CP

The expression for $H$ as a function of $F$ is $\frac{1.6}{1+8.643{\text{N}}^2/F^2}$.

Explanation of Solution

Given info: The length of the string is $80.0\text{cm}$, mass of the ball is $300\text{g}$.

The diagram is shown below.

Figure I

The formula to calculate the horizontal displacement is,

$y=\sqrt{L^2-{\left(L-H\right)}^2}$

Here,

$L$ is the length of the string.

$H$ is the height of the ball from the ground.

The formula to calculate the work done by the wind force is,

$W=Fy$

Here,

$F$ is the mass of the rider.

$y$ is the horizontal displacement.

Substitute $\sqrt{L^2-{\left(L-H\right)}^2}$ for $y$ in the above formula to find $W$.

$\begin{array}{c}W=Fy\\ =F\left(\sqrt{L^2-{\left(L-H\right)}^2}\right)\end{array}$

The formula to calculate the gravitational potential energy of the ball is,

$U=mgH$

Here,

$m$ is the mass of the ball.

$g$ is the acceleration due to gravity.

$H$ is the height of the ball from the ground.

From the law of conservation of energy,

$W=U$

Here,

$W$ is the work done by the wind force.

$U$ is the gravitational potential energy of the ball.

Substitute $F\left(\sqrt{L^2-{\left(L-H\right)}^2}\right)$ for $W$ and $mgH$ for $U$ in the above formula to find $H$.

$F\left(\sqrt{L^2-{\left(L-H\right)}^2}\right)=mgH$

Square the above expression on both sides to find $H$.

$\begin{array}{c}{F}^2\left(L^2-{\left(L-H\right)}^2\right)=m^2{g}^2{H}^2\\ F^2\left(L^2-L^2-H^2+2LH\right)=m^2{g}^2{H}^2\\ F^2\left(2LH\right)=\left(m^2{g}^2+F^2\right)H^2\\ H=\frac{F^2\left(2L\right)}{\left(m^2{g}^2+F^2\right)}\end{array}$

Substitute $300\text{g}$ for $m$, $80.0\text{cm}$ for $L$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above formula to find $H$.

$\begin{array}{c}H=\frac{F^2\left(2\left(80.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}{\left({\left(300\text{g}\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)}^2{\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}^2+F^2\right)}\\ =\frac{1.6F^2}{\left(8.643\text{N}+F^2\right)}\\ =\frac{1.6}{1+8.643{\text{N}}^2/F^2}\end{array}$

Conclusion:

Therefore, the expression for $H$ as a function of $F$ is $\frac{1.6}{1+8.643{\text{N}}^2/F^2}$.

(b)

To determine

The value of $H$ for $F=1.00\text{N}$.

Answer to Problem 8.82CP

The value of $H$ for $F=1.00\text{N}$ is $\text{0}\text{.166}\text{m}$.

Explanation of Solution

Given info: The length of the string is $80.0\text{cm}$, mass of the ball is $300\text{g}$.

From part (a), the expression for $H$ in terms of $F$ is,

$H=\frac{F^2\left(2L\right)}{\left(m^2{g}^2+F^2\right)}$

Substitute $300\text{g}$ for $m$ $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $80.0\text{cm}$ for $L$ and $1.00\text{N}$ for $F$ in the above formula to find $H$.

$\begin{array}{c}H=\frac{{\left(1.00\text{N}\right)}^2\left(2\left(80.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}{\left({\left(300\text{g}\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)}^2{\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}^2+{\left(1.00\text{N}\right)}^2\right)}\\ =\frac{1.6\text{N}\cdot \text{m}}{9.6436\text{N}}\\ =0.1659\text{m}\\ \approx \text{0}\text{.166}\text{m}\end{array}$

Conclusion:

Therefore, the value of $H$ for $F=1.00\text{N}$ is $\text{0}\text{.166}\text{m}$.

(c)

To determine

The value of $H$ for $F=10.0\text{N}$.

Answer to Problem 8.82CP

The value of $H$ for $F=10.0\text{N}$ is $1.4727\text{m}$.

Explanation of Solution

Given info: The length of the string is $80.0\text{cm}$, mass of the ball is $300\text{g}$.

From part (a), the expression for $H$ in terms of $F$ is,

$H=\frac{F^2\left(2L\right)}{\left(m^2{g}^2+F^2\right)}$

Substitute $300\text{g}$ for $m$ $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $80.0\text{cm}$ for $L$ and $10.0\text{N}$ for $F$ in the above formula to find $H$.

$\begin{array}{c}H=\frac{{\left(10.0\text{N}\right)}^2\left(2\left(80.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}{\left({\left(300\text{g}\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)}^2{\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}^2+{\left(10.0\text{N}\right)}^2\right)}\\ =\frac{160\text{N}\cdot \text{m}}{108.6436\text{N}}\\ =1.4727\text{m}\end{array}$

Conclusion:

Therefore, the value of $H$ for $F=10.0\text{N}$ is $1.4727\text{m}$.

(d)

To determine

The value of $H$ as $F$ approaches zero.

Answer to Problem 8.82CP

The value of $H$ is also approaches to zero as $F$ approaches zero.

Explanation of Solution

Given info: The length of the string is $80.0\text{cm}$, mass of the ball is $300\text{g}$.

From part (a), the expression for $H$ in terms of $F$ is,

$H=\frac{F^2\left(2L\right)}{\left(m^2{g}^2+F^2\right)}$

In the above expression, the height of the ball is directly proportional to the square of the magnitude of force as the force increases then the height of the ball also increases but in the given case, the value of force is approach to zero then the height of the ball also approach to zero.

Conclusion:

Therefore, the value of $H$ is also approaches to zero as $F$ approaches zero.

(e)

To determine

The value of $H$ as $F$ approaches infinity.

Answer to Problem 8.82CP

The value of $H$ as $F$ approaches infinity is $1.6\text{m}$.

Explanation of Solution

Given info: The length of the string is $80.0\text{cm}$, mass of the ball is $300\text{g}$.

From part (a), the expression for $H$ in terms of $F$ is,

$\begin{array}{c}H=\frac{F^2\left(2L\right)}{\left(m^2{g}^2+F^2\right)}\\ =\frac{2L}{\left(\frac{m^2{g}^2}{F^2}+1\right)}\end{array}$

Substitute $\infty$ for $F$ and $80.0\text{cm}$ for $L$ in the above formula to find $H$.

$\begin{array}{c}H=\frac{2L}{\left(\frac{m^2{g}^2}{F^2}+1\right)}\\ =\frac{2\left(80.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}{\left(\frac{m^2{g}^2}{{\left(\infty \right)}^2}+1\right)}\\ =\frac{1.6\text{m}}{\left(0+1\right)}\\ =1.6\text{m}\end{array}$

Conclusion:

Therefore, the value of $H$ as $F$ approaches infinity is $1.6\text{m}$.

(f)

To determine

The equilibrium height of the ball as a function of $F$.

Answer to Problem 8.82CP

The equilibrium height of the ball as a function of $F$ is $\left(0.08\text{m}-\frac{0.2352}{\sqrt{2.93+F^2}}\right)$

Explanation of Solution

Given info: The length of the string is $80.0\text{cm}$, mass of the ball is $300\text{g}$.

The given diagram is shown below.

Figure II

The diagram is shown below.

Figure III

From the figure the equilibrium height of the ball is,

$H_{\text{eq}}=L-L\cos \theta$

Here,

$L$ is the length of the string.

$\theta$ is the angle between the string and the vertical.

From the figure II,

$\cos \theta =\frac{mg}{\sqrt{{\left(mg\right)}^2+F^2}}$

Substitute $\frac{mg}{\sqrt{{\left(mg\right)}^2+F^2}}$ for $\cos \theta$ in formula (1) to find $H_{\text{eq}}$.

$\begin{array}{c}{H}_{\text{eq}}=L-L\cos \theta \\ =L-L\left(\frac{mg}{\sqrt{{\left(mg\right)}^2+F^2}}\right)\\ =L\left(1-\left(\frac{1}{\sqrt{1+{\left(\frac{F}{mg}\right)}^2}}\right)\right)\end{array}$

Substitute $300\text{g}$ for $m$, $80.0\text{cm}$ for $L$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above formula to find $H_{\text{eq}}$.

$\begin{array}{c}{H}_{\text{eq}}=\left(80.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(1-\left(\frac{1}{\sqrt{1+{\left(\frac{F}{\left(300\text{g}\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}\right)}^2}}\right)\right)\\ =\left(80.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(1-\frac{1}{\sqrt{1+\frac{F^2}{8.64{\text{N}}^2}}}\right)\end{array}$

Conclusion:

Therefore, the equilibrium height of the ball as a function of $F$ is $\left(80.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(1-\frac{1}{\sqrt{1+\frac{F^2}{8.64{\text{N}}^2}}}\right)$.

(g)

To determine

The value of equilibrium height of the ball $H$ for $F=10\text{N}$.

Answer to Problem 8.82CP

The value of equilibrium height of the ball $H$ for $F=10\text{N}$ is $0.5744\text{m}$.

Explanation of Solution

Given info: The length of the string is $80.0\text{cm}$, mass of the ball is $300\text{g}$.

From part (a), the expression for equivalent height of the ball in terms of $F$ is,

$H_{\text{eq}}=L-L\left(\frac{mg}{\sqrt{{\left(mg\right)}^2+F^2}}\right)$

Substitute $10\text{N}$ for $F$, $300\text{g}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $80.0\text{cm}$ for $L$ in the above formula to find $H_{\text{eq}}$.

$\begin{array}{c}{H}_{\text{eq}}=\left(80.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(1-\left(\frac{\left(300\text{g}\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}{\sqrt{{\left(\left(300\text{g}\times \frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\right)}^2+{\left(10\text{N}\right)}^2}}\right)\right)\\ =\left(80.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(1-\frac{2.940\text{N}}{\sqrt{{\left(2.940\text{N}\right)}^2+{\left(10\text{N}\right)}^2}}\right)\\ =\left(0.8\text{m}\right)\left(1-0.282\text{}\text{N}\right)\\ =0.5744\text{m}\end{array}$

Conclusion:

Therefore, the value of equilibrium height of the ball $H$ for $F=10\text{N}$ is $0.5744\text{m}$.

(h)

To determine

The value of equilibrium height of the ball $H$ as $F$ approaches infinity.

Answer to Problem 8.82CP

The value of equilibrium height of the ball $H$ as $F$ approaches infinity is $0.8\text{m}$.

Explanation of Solution

Given info: The length of the string is $80.0\text{cm}$, mass of the ball is $300\text{g}$.

From part (a), the expression for equivalent height of the ball in terms of $F$ is,

$H_{\text{eq}}=L-L\left(\frac{mg}{\sqrt{{\left(mg\right)}^2+F^2}}\right)$

Substitute $\infty$ for $F$ and $80.0\text{cm}$ for $L$ in the above formula to find $H_{\text{eq}}$.

$\begin{array}{c}{H}_{\text{eq}}=\left(80.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(1-\left(\frac{mg}{\sqrt{{\left(mg\right)}^2+{\left(\infty \right)}^2}}\right)\right)\\ =\left(80.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(1-\frac{mg}{\infty }\right)\\ =\left(80.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(1-0\right)\\ =0.8\text{m}\end{array}$

Conclusion:

Therefore, the value of equilibrium height of the ball $H$ as $F$ approaches infinity is $0.8\text{m}$.