Chapter 4, Problem 7P

Two forces ${\stackrel{\to }{F}}_1$ and ${\stackrel{\to }{F}}_2$ act on a 5.00-kg object. Taking F₁ = 20.0 N and F₂ = 15.0 N, find the accelerations of the object for the configurations of forces shown in parts (a) and (b) of Figure P4.7.

Figure P4.7

To determine

The acceleration of the object.

Answer to Problem 7P

The acceleration of the object is $\underset{\_}{5.00{\text{m/s}}^{\text{2}}\text{ at an angle 36}\text{.9}°}$.

Explanation of Solution

Write the expression for net force.

${\overrightarrow{F}}_{net}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2$        (I)

Here, ${\overrightarrow{F}}_{net}$ is the net force and ${\overrightarrow{F}}_1,{\overrightarrow{F}}_2$ are the applied forces.

Write the expression for acceleration by using equation (I).

$\begin{array}{c}\overrightarrow{a}=\frac{{\overrightarrow{F}}_{net}}{m}\\ =\frac{{\overrightarrow{F}}_1+{\overrightarrow{F}}_2}{m}\end{array}$        (II)

Here, $\overrightarrow{a}$ is the acceleration, $m$ is the mass.

Write the expression for the direction of the acceleration.

$\theta ={\tan }^{-1}\left(\frac{a_y}{a_x}\right)$        (III)

Here, $\theta$ is the angle, $a_x,a_y$ are the x and y component of acceleration.

Write the expression for the magnitude of the acceleration.

$\left|\overrightarrow{a}\right|=\sqrt{a_x^2+a_y^2}$        (IV)

Here, $\left|\overrightarrow{a}\right|$ is the magnitude of acceleration.

Conclusion:

Substitute, $\left(20.0\widehat{i}\right)\text{N}$ for ${\overrightarrow{F}}_1$, $\left(15.0\widehat{j}\right)\text{N}$ for ${\overrightarrow{F}}_2$, $5.00\text{kg}$ for $m$ in equation (II) to find $\overrightarrow{a}$.

$\begin{array}{c}\overrightarrow{a}=\frac{\left(\left(20.0\widehat{i}\right)\text{N}\right)+\left(\left(15.0\widehat{j}\right)\text{N}\right)}{\left(5.00\text{kg}\right)}\\ =\left(4.00\widehat{i}+3.00\widehat{j}\right){\text{m/s}}^{\text{2}}\end{array}$

Substitute, $4.00{\text{m/s}}^{\text{2}}$ for $a_x$, $3.00{\text{m/s}}^{\text{2}}$ for $a_y$ in equation (III) to find $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{3.00{\text{m/s}}^{\text{2}}}{4.00{\text{m/s}}^{\text{2}}}\right)\\ =36.9°\end{array}$

Substitute, $4.00{\text{m/s}}^{\text{2}}$ for $a_x$, $3.00{\text{m/s}}^{\text{2}}$ for $a_y$ in equation (IV) to find $\left|\overrightarrow{a}\right|$.

$\begin{array}{c}\left|\overrightarrow{a}\right|=\sqrt{{\left(4.00{\text{m/s}}^{\text{2}}\right)}^2+{\left(3.00{\text{m/s}}^{\text{2}}\right)}^2}\\ =5.00\text{m/s}\end{array}$

Thus, the acceleration of the object is $\underset{\_}{5.00{\text{m/s}}^{\text{2}}\text{ at an angle 36}\text{.9}°}$.

To determine

The acceleration of the object.

Answer to Problem 7P

The acceleration of the object is $\underset{\_}{6.08{\text{m/s}}^{\text{2}}\text{ at an angle 25}\text{.3}°}$.

Explanation of Solution

Write the expression for net force.

${\overrightarrow{F}}_{net}={\overrightarrow{F}}_1+\left[F_2\cos \varphi \widehat{i}+F_2\sin \varphi \widehat{j}\right]$        (V)

Here, ${\overrightarrow{F}}_{net}$ is the net force, $\varphi$ is the angle and ${\overrightarrow{F}}_1,{\overrightarrow{F}}_2$ are the applied forces.

Write the expression for acceleration by using equation (I).

$\begin{array}{c}\overrightarrow{a}=\frac{{\overrightarrow{F}}_{net}}{m}\\ =\frac{{\overrightarrow{F}}_1+\left[F_2\cos \varphi \widehat{i}+F_2\sin \varphi \widehat{j}\right]}{m}\end{array}$        (VI)

Here, $\overrightarrow{a}$ is the acceleration, $m$ is the mass.

Write the expression for the direction of the acceleration.

$\theta ={\tan }^{-1}\left(\frac{a_y}{a_x}\right)$        (VII)

Here, $\theta$ is the angle, $a_x,a_y$ are the x and y component of acceleration.

Write the expression for the magnitude of the acceleration.

$\left|\overrightarrow{a}\right|=\sqrt{a_x^2+a_y^2}$        (VIII)

Here, $\left|\overrightarrow{a}\right|$ is the magnitude of acceleration.

Conclusion:

Substitute, $\left(20.0\widehat{i}\right)\text{N}$ for ${\overrightarrow{F}}_1$, $\left(15.0\widehat{j}\right)\text{N}$ for ${\overrightarrow{F}}_2$, $5.00\text{kg}$ for $m$, $60.0°$ for $\varphi$ in equation (VI) to find $\overrightarrow{a}$.

$\begin{array}{c}\overrightarrow{a}=\frac{\left(\left(20.0\widehat{i}\right)\text{N}\right)+\left[\left(\left(15.0\cos \left(60.0\right)\widehat{i}\right)\text{N}\right)+\left(\left(15.0\sin \left(60.0\right)\widehat{j}\right)\text{N}\right)\right]}{\left(5.00\text{kg}\right)}\\ =\left(5.50\widehat{i}+2.60\widehat{j}\right){\text{m/s}}^{\text{2}}\end{array}$

Substitute, $5.50{\text{m/s}}^{\text{2}}$ for $a_x$, $2.60{\text{m/s}}^{\text{2}}$ for $a_y$ in equation (VII) to find $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{5.50{\text{m/s}}^{\text{2}}}{2.60{\text{m/s}}^{\text{2}}}\right)\\ =25.3°\end{array}$

Substitute, $5.50{\text{m/s}}^{\text{2}}$ for $a_x$, $2.60{\text{m/s}}^{\text{2}}$ for $a_y$ in equation (VIII) to find $\left|\overrightarrow{a}\right|$.

$\begin{array}{c}\left|\overrightarrow{a}\right|=\sqrt{{\left(5.50{\text{m/s}}^{\text{2}}\right)}^2+{\left(2.60{\text{m/s}}^{\text{2}}\right)}^2}\\ =6.08{\text{m/s}}^{\text{2}}\end{array}$

Thus, the acceleration of the object is $\underset{\_}{6.08{\text{m/s}}^{\text{2}}\text{ at an angle 25}\text{.3}°}$.