Chapter 4, Problem 50P

(a)

To determine

The expression for the tension in each section of string.

Answer to Problem 50P

The expression for the tension in each section of string is $T_1=\frac{2mg}{\sin {\theta }_1},T_2=\frac{mg}{\sin \left[{\tan }^{-1}\left(\frac{1}{2}\tan {\theta }_1\right)\right]},T_3=\frac{2mg}{\tan {\theta }_1}$.

Explanation of Solution

The free body diagram for the system is given by figure 1.

Write the expression for the net force from Newton’s second law.

$T_2\cos {\theta }_2-T_1\cos {\theta }_1=0$        (I)

$T_1\sin {\theta }_1-T_2\sin {\theta }_2-mg=0$        (II)

$T_2\cos {\theta }_2-T_3=0$        (III)

$T_2\sin {\theta }_2-mg=0$        (IV)

Here, $T_1,T_2,T_3$ are the tensions, ${\theta }_1,{\theta }_2$ are the angles, $m$ is the mass and $g$ is the gravitational acceleration.

Conclusion:

Substitute, $mg$ for $T_2\sin {\theta }_2$ in Equation (II) to find $T_1$.

  $\begin{array}{l}{T}_1\sin {\theta }_1-mg-mg=0\\ T_1=\frac{2mg}{\sin {\theta }_1}\end{array}$

Substitute, $T_3$ for $T_2\cos {\theta }_2$, $\left(\frac{2mg}{\sin {\theta }_1}\right)$ for $T_1$ in Equation (I) to find $T_3$.

$\begin{array}{l}{T}_3-\left(\frac{2mg}{\sin {\theta }_1}\right)\cos {\theta }_1=0\\ T_3=\frac{2mg}{\tan {\theta }_1}\end{array}$

Divide equation (IV) by (III) and substitute $\left(\frac{2mg}{\tan {\theta }_1}\right)$ for $T_3$ to find ${\theta }_2$.

$\begin{array}{l}\frac{T_2\sin {\theta }_2}{T_2\cos {\theta }_2}=\frac{mg}{T_3}\\ \frac{T_2\sin {\theta }_2}{T_2\cos {\theta }_2}=\frac{mg}{\left(\frac{2mg}{\tan {\theta }_1}\right)}\\ {\theta }_2={\tan }^{-1}\left(\frac{\tan {\theta }_1}{2}\right)\end{array}$

Substitute, ${\tan }^{-1}\left(\frac{\tan {\theta }_1}{2}\right)$ for ${\theta }_2$ in Equation (IV) to find $T_2$.

$\begin{array}{l}{T}_2\sin \left[{\tan }^{-1}\left(\frac{\tan {\theta }_1}{2}\right)\right]-mg=0\\ T_2=\frac{mg}{\sin \left[{\tan }^{-1}\left(\frac{\tan {\theta }_1}{2}\right)\right]}\end{array}$

Thus, the expression for the tension in each section of string is $T_1=\frac{2mg}{\sin {\theta }_1},T_2=\frac{mg}{\sin \left[{\tan }^{-1}\left(\frac{1}{2}\tan {\theta }_1\right)\right]},T_3=\frac{2mg}{\tan {\theta }_1}$.

(b)

To determine

The expression for the angle ${\theta }_2$ in terms of ${\theta }_1$.

Answer to Problem 50P

The expression for the angle ${\theta }_2$ in terms of ${\theta }_1$ is $\underset{\_}{{\theta }_2={\tan }^{-1}\left(\frac{\tan {\theta }_1}{2}\right)}$.

Explanation of Solution

Write the expression for the net force from Newton’s second law.

$T_2\cos {\theta }_2-T_1\cos {\theta }_1=0$        (I)

$T_1\sin {\theta }_1-T_2\sin {\theta }_2-mg=0$        (II)

$T_2\cos {\theta }_2-T_3=0$        (III)

$T_2\sin {\theta }_2-mg=0$        (IV)

Conclusion:

Divide equation (IV) by (III) and substitute $\left(\frac{2mg}{\tan {\theta }_1}\right)$ for $T_3$ to find ${\theta }_2$.

$\begin{array}{l}\frac{T_2\sin {\theta }_2}{T_2\cos {\theta }_2}=\frac{mg}{T_3}\\ \frac{T_2\sin {\theta }_2}{T_2\cos {\theta }_2}=\frac{mg}{\left(\frac{2mg}{\tan {\theta }_1}\right)}\\ {\theta }_2={\tan }^{-1}\left(\frac{\tan {\theta }_1}{2}\right)\end{array}$

Thus, the expression for the angle ${\theta }_2$ in terms of ${\theta }_1$ is $\underset{\_}{{\theta }_2={\tan }^{-1}\left(\frac{\tan {\theta }_1}{2}\right)}$.

(c)

To determine

The expression for the distance between the endpoints of the string.

Answer to Problem 50P

The expression for the distance between the endpoints of the string is $\underset{\_}{D=\frac{L}{5}\left[\left(2\cos {\theta }_1\right)+\left[2\cos \left\{{\tan }^{-1}\left(\frac{1}{2}\tan {\theta }_1\right)\right\}\right]+1\right]}$.

Explanation of Solution

Write the expression for the total length.

$L=5l$        (V)

Here, $L$ is the total length and $l$ is the length of each portion.

Write the expression for the distance.

$D=2l\cos {\theta }_1+2l\cos {\theta }_2+l$        (VI)

Conclusion:

Substitute, $\left(L/5\right)$ for $l$, ${\tan }^{-1}\left(\frac{\tan {\theta }_1}{2}\right)$ for ${\theta }_2$ in Equation (VI) to find $D$.

  $\begin{array}{c}D=2\left(L/5\right)\cos {\theta }_1+2\left(L/5\right)\cos \left[{\tan }^{-1}\left(\frac{\tan {\theta }_1}{2}\right)\right]+\left(L/5\right)\\ =\frac{L}{5}\left[\left(2\cos {\theta }_1\right)+\left[2\cos \left\{{\tan }^{-1}\left(\frac{1}{2}\tan {\theta }_1\right)\right\}\right]+1\right]\end{array}$

Thus, the expression for the distance between the endpoints of the string is $\underset{\_}{D=\frac{L}{5}\left[\left(2\cos {\theta }_1\right)+\left[2\cos \left\{{\tan }^{-1}\left(\frac{1}{2}\tan {\theta }_1\right)\right\}\right]+1\right]}$.