Chapter 2, Problem 35P

(a)

To determine

The angle between the two vectors if the net force is $7.0\text{ N}$.

Answer to Problem 35P

The angle between the two vectors is $0°$.

Explanation of Solution

Write the expression for the magnitude of the sum of two forces.

    $\begin{array}{c}{\left|\overrightarrow{A}+\overrightarrow{B}\right|}^2=A^2+B^2+2AB\cos \theta \\ \theta ={\cos }^{-1}\left(\frac{{\left|\overrightarrow{A}+\overrightarrow{B}\right|}^2-A^2-B^2}{2AB}\right)\end{array}$

Here, $\overrightarrow{A}$, $\overrightarrow{B}$ are two vectors, $A$, $B$ are their magnitude and $\theta$ is the angle between them.

Conclusion:

Substitute $3.0\text{N}$ for $A$, $4.0\text{N}$ for $B$ and $7.0\text{ N}$ for $\left|\overrightarrow{A}+\overrightarrow{B}\right|$ in the above equation to find $\theta$

    $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{{\left(7.0\text{N}\right)}^2-{\left(3.0\text{N}\right)}^2-{\left(4.0\text{N}\right)}^2}{2\left(3.0\text{N}\right)\left(4.0\text{N}\right)}\right)\\ ={\cos }^{-1}\left(1\right)\\ =0°\end{array}$

Therefore, the angle between the two vectors is $0°$.

(b)

To determine

The angle between the two vectors if the net force is $5.0\text{N}$ .

Answer to Problem 35P

The angle between the two vectors is $90°$.

Explanation of Solution

Write the expression for the magnitude of the sum of two forces.

    $\theta ={\cos }^{-1}\left(\frac{{\left|\overrightarrow{A}+\overrightarrow{B}\right|}^2-A^2-B^2}{2AB}\right)$

Conclusion:

Substitute $3.0\text{N}$ for $A$, $4.0\text{N}$ for $B$ and $5.0\text{ N}$ for $\left|\overrightarrow{A}+\overrightarrow{B}\right|$ in the above equation to find $\theta$

    $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{{\left(5.0\text{N}\right)}^2-{\left(3.0\text{N}\right)}^2-{\left(4.0\text{N}\right)}^2}{2\left(3.0\text{N}\right)\left(4.0\text{N}\right)}\right)\\ ={\cos }^{-1}\left(0\right)\\ =90°\end{array}$

Therefore, the angle between the two vectors is $90°$.

(c)

To determine

The angle that gives the smallest magnitude of the net force and also the magnitude.

Answer to Problem 35P

The angle that gives the smallest magnitude of the net force is $180°$ and also the magnitude is $1.0\text{N}$.

Explanation of Solution

For, the smallest magnitude, both vectors should be in the opposite direction.

Recall the expression for magnitude of net force from part (a)

    $\left|\overrightarrow{A}+\overrightarrow{B}\right|=\sqrt{A^2+B^2+2AB\cos \theta }$

Here, $\overrightarrow{A}$, $\overrightarrow{B}$ are two vectors, $A$, $B$ are their magnitude and $\theta$ is the angle between them.

Conclusion:

Substitute $3.0\text{N}$ for $A$, $4.0\text{N}$ for $B$ and $180°$ for $\left|\overrightarrow{A}+\overrightarrow{B}\right|$ in the above equation to find $\left|\overrightarrow{A}+\overrightarrow{B}\right|$

    $\begin{array}{c}\left|\overrightarrow{A}+\overrightarrow{B}\right|=\sqrt{{\left(3.0\text{N}\right)}^2+{\left(4.0\text{N}\right)}^2+2\left(3.0\text{N}\right)\left(4.0\text{N}\right)\left(-1\right)}\\ =1.0\text{N}\end{array}$

Therefore, the angle that gives the smallest magnitude of the net force is $180°$ and also the magnitude is $1.0\text{N}$.