Chapter 11, Problem 11.6P

(a)

To determine

The quantity ${\cos }^{-1}\left(\frac{\overrightarrow{\text{A}}\cdot \overrightarrow{\text{B}}}{\text{AB}}\right)$.

Answer to Problem 11.6P

The quantity ${\cos }^{-1}\left(\frac{\overrightarrow{\text{A}}\cdot \overrightarrow{\text{B}}}{\text{AB}}\right)$ is $168°$.

Explanation of Solution

Given info: The vector $\overrightarrow{\text{B}}$ is $6\widehat{\text{i}}-10\widehat{\text{j}}+9\widehat{\text{k}}$ and vector $\overrightarrow{\text{A}}$ is $-3\widehat{\text{i}}+7\widehat{\text{j}}-4\widehat{\text{k}}$

The formula to calculate the dot product of two vector is,

$\begin{array}{c}\overrightarrow{\text{A}}\cdot \overrightarrow{\text{B}}=\left(x_1\widehat{\text{i}}+x_2\widehat{\text{j}}+x_3\widehat{\text{k}}\right)\left(y_1\widehat{\text{i}}+y_2\widehat{\text{j}}+y_3\widehat{\text{k}}\right)\\ =\left(x_1{y}_1+x_2{y}_2+x_3{y}_3\right)\end{array}$

Here,

$x_1$ is the $x$-component of first vector.

$x_2$ is the $y$-component of first vector.

$x_3$ is the $z$-component of first vector.

$y_1$ is the $x$ component of second vector.

$y_2$ is the $y$-component of second vector.

$y_3$ is the $z$- component of second vector.

Substitute $-3$ for $x_1$,$7$ for $x_2$,$-4$ for $x_3$,$6$ for $y_1$,$-10$ for $y_2$,$9$ for $y_3$ in the above formula to find $\overrightarrow{\text{A}}\cdot \overrightarrow{\text{B}}$.

$\begin{array}{c}\overrightarrow{\text{A}}\cdot \overrightarrow{\text{B}}=\left(-3\right)\left(6\right)+\left(7\right)\left(-10\right)+\left(-4\right)\left(9\right)\\ =-18-70-36\\ =-124\end{array}$

Thus, the value of dot product of vectors is $-124$.

The formula to calculate the magnitude of vector $\overrightarrow{\text{A}}$ is,

$\left|\overrightarrow{\text{A}}\right|=\sqrt{{\left(x_1\right)}^2+{\left(x_2\right)}^2+{\left(x_3\right)}^2}$

Here,

$x_1$ is the $x$-component of first vector.

$x_2$ is the $y$-component of first vector.

$x_3$ is the $z$-component of first vector.

Substitute $-3$ for $x_1$,$7$ for $x_2$,$-4$ for $x_3$ in the above formula to find $\left|\overrightarrow{\text{A}}\right|$

$\begin{array}{c}\left|\overrightarrow{\text{A}}\right|=\sqrt{{\left(-3\right)}^2+{\left(7\right)}^2+{\left(-4\right)}^2}\\ =\sqrt{9+49+16}\\ =\sqrt{74}\\ =8.60\end{array}$

Thus, the value of magnitude of vector $\overrightarrow{\text{A}}$ is $8.60$.

The formula to calculate the magnitude of vector $\overrightarrow{\text{B}}$ is,

$\left|\overrightarrow{\text{B}}\right|=\sqrt{{\left(y_1\right)}^2+{\left(y_2\right)}^2+{\left(y_3\right)}^2}$

$y_1$ is the $x$ component of second vector.

$y_2$ is the $y$-component of second vector.

$y_3$ is the $z$- component of second vector.

Substitute $6$ for $y_1$,$-10$ for $y_2$,$9$ for $y_3$ in the above formula to find $\left|\overrightarrow{\text{B}}\right|$

$\begin{array}{c}\left|\overrightarrow{\text{B}}\right|=\sqrt{{\left(6\right)}^2+{\left(-10\right)}^2+{\left(9\right)}^2}\\ =\sqrt{36+100+81}\\ =\sqrt{217}\\ =14.73\end{array}$

Thus, the value of magnitude of vector $\overrightarrow{\text{B}}$ is $14.7$.

The formula to calculate the angle between the vectors is,

$\begin{array}{c}\cos \theta =\frac{\overrightarrow{\text{A}}\cdot \overrightarrow{\text{B}}}{\left|\overrightarrow{\text{A}}\right|\left|\overrightarrow{\text{B}}\right|}\\ \theta ={\cos }^{-1}\left(\frac{\overrightarrow{\text{A}}\cdot \overrightarrow{\text{B}}}{\left|\overrightarrow{\text{A}}\right|\left|\overrightarrow{\text{B}}\right|}\right)\end{array}$

Here,

$\overrightarrow{\text{A}}\cdot \overrightarrow{\text{B}}$ is the dot product of  two vectors.

$\left|\overrightarrow{\text{A}}\right|\left|\overrightarrow{\text{B}}\right|$ is the magnitude of two vectors.

Substitute $-124$ for $\overrightarrow{\text{A}}\cdot \overrightarrow{\text{B}}$,$8.60$ for $\left|\overrightarrow{\text{A}}\right|$,$14.7$ for $\left|\overrightarrow{\text{B}}\right|$ in the above formula to find $\theta$.

$\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{\overrightarrow{\text{A}}\cdot \overrightarrow{\text{B}}}{\left|\overrightarrow{\text{A}}\right|\left|\overrightarrow{\text{B}}\right|}\right)\\ ={\cos }^{-1}\left(\frac{-124}{\left(8.60\right)\left(14.7\right)}\right)\\ =168.52°\\ \approx 168°\end{array}$

Conclusion:

Therefore, the quantity ${\cos }^{-1}\left(\frac{\overrightarrow{\text{A}}\cdot \overrightarrow{\text{B}}}{\text{AB}}\right)$ is $168°$

(b)

To determine

The magnitude of the total angular momentum of the system of the system about the axle of the pulley.

Answer to Problem 11.6P

The magnitude of the total angular momentum of the system of the system about the axle of the pulley is $11.88°$.

Explanation of Solution

Given info: The mass of counterweight is $4.00\text{kg}$,mass of pulley is $2.00\text{kg}$,radius of the pulley is $8.00\text{cm}$.

The formula to calculate the cross product of two vector is,

$\overrightarrow{\text{A}}\times \overrightarrow{\text{B}}=\left|\begin{array}{ccc}\widehat{\text{i}}& \widehat{\text{j}}& \widehat{\text{k}}\\ x_1& x_2& x_3\\ y_1& y_2& y_3\end{array}\right|$

Substitute $-3$ for $x_1$,$7$ for $x_2$,$-4$ for $x_3$,$6$ for $y_1$,$-10$ for $y_2$,$9$ for $y_3$ in the above formula to find $\overrightarrow{\text{A}}\times \overrightarrow{\text{B}}$.

$\begin{array}{c}\overrightarrow{\text{A}}\times \overrightarrow{\text{B}}=\left|\begin{array}{ccc}\widehat{\text{i}}& \widehat{\text{j}}& \widehat{\text{k}}\\ -3& 7& -4\\ 6& -10& 9\end{array}\right|\\ =\widehat{\text{i}}\left(7\times 9-\left(-4\times -9\right)\right)-\widehat{\text{j}}\left(-3\times 9-\left(-4\times 6\right)\right)+\widehat{\text{k}}\left(-3\times -10\right)-\left(7\times 6\right)\\ =23\widehat{\text{i}}+3\widehat{\text{j}}-12\widehat{\text{k}}\end{array}$

Thus, the value of cross product of vectors is $23\widehat{\text{i}}+3\widehat{\text{j}}-12\widehat{\text{k}}$.

The formula to calculate the magnitude of cross product of two vectors is

$\left|\overrightarrow{\text{A}}\times \overrightarrow{\text{B}}\right|=\sqrt{{\left(a_1\right)}^2+{\left(a_2\right)}^2+{\left(a_3\right)}^2}$

Here,

$a_1$ is the $x$-component of $\left|\overrightarrow{\text{A}}\times \overrightarrow{\text{B}}\right|$

$a_2$ is the $y$-component of $\left|\overrightarrow{\text{A}}\times \overrightarrow{\text{B}}\right|$

$a_3$ s the $z$-component of $\left|\overrightarrow{\text{A}}\times \overrightarrow{\text{B}}\right|$

Substitute $23$ for $a_1$, $3$ for $a_2$ and $-12$ for $a_3$ in the above formula to find $\left|\overrightarrow{\text{A}}\times \overrightarrow{\text{B}}\right|$.

$\begin{array}{c}\left|\overrightarrow{\text{A}}\times \overrightarrow{\text{B}}\right|=\sqrt{{\left(23\right)}^2+{\left(3\right)}^2+{\left(-12\right)}^2}\\ =\sqrt{529+9+144}\\ =\sqrt{682}\\ =26.115\end{array}$

Thus, the value of magnitude of $\left|\overrightarrow{\text{A}}\times \overrightarrow{\text{B}}\right|$ is $26.115$.

The formula to calculate the magnitude of vector $\overrightarrow{A}$ is,

$\left|\overrightarrow{\text{A}}\right|=\sqrt{{\left(x_1\right)}^2+{\left(x_2\right)}^2+{\left(x_3\right)}^2}$

Here,

$x_1$ is the $x$-component of first vector.

$x_2$ is the $y$-component of first vector.

$x_3$ is the $z$-component of first vector.

Substitute $-3$ for $x_1$,$7$ for $x_2$,$-4$ for $x_3$ in the above formula to find $\left|\overrightarrow{\text{A}}\right|$

$\begin{array}{c}\left|\overrightarrow{\text{A}}\right|=\sqrt{{\left(-3\right)}^2+{\left(7\right)}^2+{\left(-4\right)}^2}\\ =\sqrt{9+49+16}\\ =\sqrt{74}\\ =8.60\end{array}$

Thus, the value of magnitude of vector $\overrightarrow{\text{A}}$ is $8.60$.

The formula to calculate the magnitude of vector $\overrightarrow{\text{B}}$ is,

$\left|\overrightarrow{\text{B}}\right|=\sqrt{{\left(y_1\right)}^2+{\left(y_2\right)}^2+{\left(y_3\right)}^2}$

$y_1$ is the $x$ component of second vector.

$y_2$ is the $y$-component of second vector.

$y_3$ is the $z$- component of second vector.

Substitute $6$ for $y_1$,$-10$ for $y_2$,$9$ for $y_3$ in the above formula to find $\left|\overrightarrow{\text{B}}\right|$

$\begin{array}{c}\left|\overrightarrow{\text{B}}\right|=\sqrt{{\left(6\right)}^2+{\left(-10\right)}^2+{\left(9\right)}^2}\\ =\sqrt{36+100+81}\\ =\sqrt{217}\\ =14.73\end{array}$

Thus, the value of magnitude of vector $\overrightarrow{\text{B}}$ is $14.7$.

The formula to calculate the angle between the vectors is,

$\begin{array}{c}\sin \theta =\frac{\left|\overrightarrow{\text{A}}\times \overrightarrow{\text{B}}\right|}{\left|\overrightarrow{\text{A}}\right|\left|\overrightarrow{\text{B}}\right|}\\ \theta ={\sin }^{-1}\left(\frac{\overrightarrow{\text{A}}\times \overrightarrow{\text{B}}}{\left|\overrightarrow{\text{A}}\right|\left|\overrightarrow{\text{B}}\right|}\right)\end{array}$

Here,

$\left|\overrightarrow{\text{A}}\times \overrightarrow{\text{B}}\right|$ is the magnitude of cross  product of  two vectors.

$\left|\overrightarrow{\text{A}}\right|\left|\overrightarrow{\text{B}}\right|$ is the magnitude of two vectors.

Substitute $26.115$ for $\overrightarrow{\text{A}}\times \overrightarrow{\text{B}}$,$8.60$ for $\left|\overrightarrow{\text{A}}\right|$,$14.7$ for $\left|\overrightarrow{\text{B}}\right|$ in the above formula to find $\theta$.

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{26.115}{\left(8.60\right)\left(14.7\right)}\right)\\ ={\sin }^{-1}\left(\frac{26.115}{126.42}\right)\\ ={\sin }^{-1}\left(0.206\right)\\ =11.88°\end{array}$

Conclusion:

Therefore, the quantity ${\sin }^{-1}\left(\frac{\overrightarrow{\text{A}}\times \overrightarrow{\text{B}}}{\left|\overrightarrow{\text{A}}\right|\left|\overrightarrow{\text{B}}\right|}\right)$ is $11.88°$

(c)

To determine

Whether the angle between the vectors is given buy part (a) or part (b).

Answer to Problem 11.6P

The angle between the vectors is given by part (a).

Explanation of Solution

Given info: The vector $\overrightarrow{B}$ is $6\widehat{\text{i}}-10\widehat{\text{j}}+9\widehat{\text{k}}$ and vector $\overrightarrow{A}$ is $-3\widehat{\text{i}}+7\widehat{\text{j}}-4\widehat{\text{k}}$

From part (a), the value of the angle obtained is $168°$ and the value of the angle obtained from part (b) is $11.88°$. The value of angle obtained in part (a) is same as obtained in part (b) but the difference is the value of angle obtained in part (a) is anticlockwise.

The value of angle obtained in part (a) is,

$\begin{array}{c}\sin \left(168°\right)=\sin \left(180°-168°\right)\\ =-\cos \left(11.88°\right)\\ =\cos \left(11.88°\right)\end{array}$

Conclusion:

Therefore, the angle between the vectors is given by part (a).