Chapter 5, Problem 5.19P

Two forces ${\stackrel{\to }{\text{F}}}_1$ and ${\stackrel{\to }{\text{F}}}_2$ 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 P5.19.

Figure P5.19

To determine

The acceleration of the object for the configurations of forces shown in part (a) of given figure.

Answer to Problem 5.19P

The magnitude of the acceleration is $5\text{m}/{\text{s}}^2$ and the angle made by the acceleration with the horizontal is $36.869°$.

Explanation of Solution

Given info: Two forces ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$ act on a object which mass is $5.00\text{kg}$. The value of ${\overrightarrow{F}}_1$ is $20.0\text{N}$ and the value of ${\overrightarrow{F}}_2$ is $15.0\text{N}$ and the angle between ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$ is $90°$.

The net force act on the object is,

$\overrightarrow{F}={\overrightarrow{F}}_{\text{x}}\widehat{i}+{\overrightarrow{F}}_{\text{y}}\widehat{j}$ (1)

Here,

$F_{\text{x}}$ is the horizontal component of force.

$F_{\text{y}}$ is the vertical component of force.

The horizontal component of force is,

$F_{\text{x}}=F_1\cos {\theta }_1+F_2\cos {\theta }_2$

In the horizontal direction the ${\overrightarrow{F}}_1$ makes an angle $0°$ and ${\overrightarrow{F}}_2$ makes an angle $90°$.

Substitute $0°$ for ${\theta }_1$, $90°$ for ${\theta }_2$, $20.0\text{N}$ for ${\overrightarrow{F}}_1$ and $15.0\text{N}$ for ${\overrightarrow{F}}_2$ in above equation.

$\begin{array}{c}{F}_{\text{x}}=\left(20.0\text{N}\right)\cos 0°+\left(15.0\text{N}\right)\cos 90°\\ =20.0\text{N}+0\\ =20.0\text{N}\end{array}$

The vertical component of force is,

$F_{\text{y}}=F_1\sin {\theta }_1+F_2\sin {\theta }_2$

In the vertical direction the ${\overrightarrow{F}}_1$ makes an angle $90°$ and ${\overrightarrow{F}}_2$ makes an angle $0°$.

Substitute $90°$ for ${\theta }_1$, $0°$ for ${\theta }_2$, $20.0\text{N}$ for ${\overrightarrow{F}}_1$ and $15.0\text{N}$ for ${\overrightarrow{F}}_2$ in above equation.

$\begin{array}{c}{F}_{\text{y}}=\left(20.0\text{N}\right)\sin 0°+\left(15.0\text{N}\right)\sin 90°\\ =0+15.0\text{N}\\ =15.0\text{N}\end{array}$

Substitute $20.0\text{N}$ for $F_{\text{x}}$ and $15.0\text{N}$ for $F_{\text{y}}$ in equation (1).

$\overrightarrow{F}=\left(20.0\widehat{i}+15.0\widehat{j}\right)\text{N}$

According to Newton’s second law of motion,

$\overrightarrow{F}=m\overrightarrow{a}$

Here,

$\overrightarrow{a}$ is the acceleration of the object.

Substitute $\left(20.0\widehat{i}+15.0\widehat{j}\right)\text{N}$ for $\overrightarrow{F}$ and $5.00\text{kg}$ for $m$ in above equation.

$\begin{array}{c}\left(20.0\widehat{i}+15.0\widehat{j}\right)\text{N}=\left(5.00\text{kg}\right)\overrightarrow{a}\\ \overrightarrow{a}=\frac{\left(20.0\widehat{i}+15.0\widehat{j}\right)\text{N}}{\left(5.00\text{kg}\right)}\\ =\left(4\widehat{i}+3\widehat{j}\right)\text{m}/{\text{s}}^2\end{array}$

The magnitude of the acceleration is,

$a=\sqrt{a_{\text{x}}{}^2+a_{\text{y}}{}^2}$

Here,

$a_{\text{x}}$ is the acceleration in horizontal direction.

$a_{\text{y}}$ is the acceleration in vertical direction.

Substitute $4$ for $a_{\text{x}}$ and $3$ for $a_{\text{y}}$ in above equation.

$\begin{array}{c}a=\sqrt{4^2+3^2}\text{m}/{\text{s}}^2\\ =\left(\sqrt{16+9}\right)\text{m}/{\text{s}}^2\\ =\sqrt{25}\text{m}/{\text{s}}^2\\ =5.00\text{m}/{\text{s}}^2\end{array}$

The angle made by acceleration with the horizontal is,

$\tan \theta =\frac{a_{\text{y}}}{a_{\text{x}}}$

Substitute $4$ for $a_{\text{x}}$ and $3$ for $a_{\text{y}}$ in above equation.

$\begin{array}{c}\tan \theta =\frac{3}{4}\\ \theta ={\tan }^{-1}\left(\frac{3}{4}\right)\\ =36.869°\end{array}$

Conclusion:

Therefore, the magnitude of the acceleration is $5.00\text{m}/{\text{s}}^2$ and the angle made by the acceleration with the horizontal is $36.869°$.

To determine

The acceleration of the object for the configurations of forces shown in part (b) of given figure.

Answer to Problem 5.19P

The magnitude of the acceleration is $6.08{\text{m/s}}^2$ and the angle made by the acceleration with the horizontal is $25.3°$.

Explanation of Solution

Given info: Two forces ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$ act on a object which mass is $5.00\text{kg}$. The value of ${\overrightarrow{F}}_1$ is $20.0\text{N}$ and the value of ${\overrightarrow{F}}_2$ is $15.0\text{N}$ and the angle between ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$ is $60°$.

The horizontal component of force is,

$F_{\text{x}}=F_1\cos {\theta }_1+F_2\cos {\theta }_2$

In the horizontal direction the ${\overrightarrow{F}}_1$ makes an angle $0°$ and ${\overrightarrow{F}}_2$ makes an angle $60°$.

Substitute $0°$ for ${\theta }_1$, $60°$ for ${\theta }_2$, $20.0\text{N}$ for ${\overrightarrow{F}}_1$ and $15.0\text{N}$ for ${\overrightarrow{F}}_2$ in above equation.

$\begin{array}{c}{F}_{\text{x}}=\left(20.0\text{N}\right)\cos 0°+\left(15.0\text{N}\right)\cos 60°\\ =20.0\text{N}+\left(15.0\text{N}\right)\left(0.5\right)\\ =27.5\text{N}\end{array}$

The vertical component of force is,

$F_{\text{y}}=F_1\sin {\theta }_1+F_2\sin {\theta }_2$

In the vertical direction the ${\overrightarrow{F}}_1$ makes an angle $60°$ and ${\overrightarrow{F}}_2$ makes an angle $0°$.

Substitute $60°$ for ${\theta }_1$, $0°$ for ${\theta }_2$, $20.0\text{N}$ for ${\overrightarrow{F}}_1$ and $15.0\text{N}$ for ${\overrightarrow{F}}_2$ in above equation.

$\begin{array}{c}{F}_{\text{y}}=\left(20.0\text{N}\right)\sin 0°+\left(15.0\text{N}\right)\sin 60°\\ =0+\left(15.0\text{N}\right)\left(0.866\right)\\ =13.0\text{N}\end{array}$

Substitute $27.5\text{N}$ for $F_{\text{x}}$ and $13.0\text{N}$ for $F_{\text{y}}$ in equation (1).

$\overrightarrow{F}=\left(27.5\widehat{i}+13.0\widehat{j}\right)\text{N}$

According to Newton’s second law of motion,

$\overrightarrow{F}=m\overrightarrow{a}$

Substitute $\left(27.5\widehat{i}+13.0\widehat{j}\right)\text{N}$ for $\overrightarrow{F}$ and $5.00\text{kg}$ for $m$ in above equation.

$\begin{array}{c}\left(27.5\widehat{i}+13.0\widehat{j}\right)\text{N}=\left(5.00\text{kg}\right)\overrightarrow{a}\\ \overrightarrow{a}=\frac{\left(27.5\widehat{i}+13.0\widehat{j}\right)\text{N}}{\left(5.00\text{kg}\right)}\\ =\left(5.5\widehat{i}+2.6\widehat{j}\right)\text{m}/{\text{s}}^2\end{array}$

The magnitude of the acceleration is,

$a=\sqrt{a_{\text{x}}{}^2+a_{\text{y}}{}^2}$

Substitute $5.5$ for $a_{\text{x}}$ and $2.6$ for $a_{\text{y}}$ in above equation.

$\begin{array}{c}a=\sqrt{{\left(5.5\right)}^2+{\left(2.6\right)}^2}\text{m}/{\text{s}}^2\\ =\left(\sqrt{30.25+6.76}\right)\text{m}/{\text{s}}^2\\ =\sqrt{37.01}\text{m}/{\text{s}}^2\\ =6.08\text{m}/{\text{s}}^2\end{array}$

The angle made by acceleration with the horizontal is,

$\tan \theta =\frac{a_{\text{y}}}{a_{\text{x}}}$

Substitute $5.5$ for $a_{\text{x}}$ and $2.6$ for $a_{\text{y}}$ in above equation.

$\begin{array}{c}\tan \theta =\frac{5.5}{2.6}\\ \theta ={\tan }^{-1}\left(\frac{5.5}{2.6}\right)\\ ={\tan }^{-1}\left(2.115\right)\\ =25.3°\end{array}$

Conclusion:

Therefore, the magnitude of the acceleration is $6.08\text{m}/{\text{s}}^2$ and the angle made by the acceleration with the horizontal is $25.3°$.