Chapter 5, Problem 19P

(a)

To determine

The acceleration of the object.

Answer to Problem 19P

The acceleration of the object is $5.00\text{m}/{\text{s}}^{\text{2}}$ at an angle of $36.9°$ from horizontal.

Explanation of Solution

The forces on the object are acting in the two direction so the acceleration of the object will be at some from the horizontal.

Write the expression for the acceleration of the object in horizontal direction as.

  $a_h=\frac{F_1}{m}$                                                                                                           (I)

Here, $a_h$ is the acceleration in horizontal direction, $F_1$ is the force in horizontal direction and $m$ is the mass of the object.

Write the expression for the acceleration of the object in vertical direction as.

  $a_v=\frac{F_2}{m}$                                                                                                         (II)

Here, $a_v$ is the acceleration in vertical direction and $F_2$ is the force in vertical direction.

Write the expression for the net acceleration of the object as.

  $a=\sqrt{{\left(a_h\right)}^2+{\left(a_v\right)}^2+2a_h{a}_v\cos \varphi }$                                                                 (III)

Here, $a$ is the net acceleration of the object and $\varphi$ is the angle between two forces.

Write the expression for the angle of the net acceleration as.

  $\tan {\theta }_1=\frac{a_v}{a_h}$                                                                                                  (IV)

Here, ${\theta }_1$ is the angle made by the net acceleration from the horizontal.

Conclusion:

Substitute $20.0\text{N}$ for $F_1$ and $5.00\text{kg}$ for $m$ in equation (I).

  $\begin{array}{c}{a}_h=\frac{20.0\text{N}}{5.00\text{kg}}\\ =4.00\text{m}/{\text{s}}^{\text{2}}\end{array}$

Substitute $15.0\text{N}$ for $F_2$ and $5.00\text{kg}$ for $m$ in equation (II).

  $\begin{array}{c}{a}_v=\frac{15.0\text{N}}{5.00\text{kg}}\\ =3.00\text{m}/{\text{s}}^{\text{2}}\end{array}$

Substitute $4.00\text{m}/{\text{s}}^{\text{2}}$ for $a_h$, $3.00\text{m}/{\text{s}}^{\text{2}}$ for $a_v$ and $90.0°$ for $\varphi$ in equation (III).

  $\begin{array}{c}a=\sqrt{{\left(4.00\text{m}/{\text{s}}^{\text{2}}\right)}^2+{\left(3.00\text{m}/{\text{s}}^{\text{2}}\right)}^2+2\left(4.00\text{m}/{\text{s}}^{\text{2}}\right)\left(3.00\text{m}/{\text{s}}^{\text{2}}\right)\left(\cos 90.0°\right)}\\ =\left(\sqrt{16.00+9.00}\right)\text{m}/{\text{s}}^{\text{2}}\\ =5.00\text{m}/{\text{s}}^{\text{2}}\end{array}$

Substitute $4.00\text{m}/{\text{s}}^{\text{2}}$ for $a_h$ and $3.00\text{m}/{\text{s}}^{\text{2}}$ for $a_v$ in equation (IV).

  $\begin{array}{c}\tan {\theta }_1=\frac{\left(3.00\text{m}/{\text{s}}^{\text{2}}\right)}{\left(4.00\text{m}/{\text{s}}^{\text{2}}\right)}\\ {\theta }_1={\tan }^{-1}\left(\frac{3}{4}\right)\\ =36.86°\\ \approx 36.9°\end{array}$

Thus, the acceleration of the object is $5.00\text{m}/{\text{s}}^{\text{2}}$ at an angle of $36.9°$ from horizontal.

(b)

To determine

The acceleration of the object.

Answer to Problem 19P

The acceleration of the object is $6.08\text{m}/{\text{s}}^{\text{2}}$ at an angle of $25.3°$ from horizontal.

Explanation of Solution

Write the expression for the angle made by the net acceleration as.

  $\tan {\theta }_2=\frac{a_v\sin {\varphi }_2}{a_h+a_v\cos {\varphi }_2}$                                                                               (V)

Here, ${\theta }_2$ is the angle made by the net acceleration from horizontal and ${\varphi }_2$ is the angle between $a_h$ and $a_v$ .

Conclusion:

Substitute $4.00\text{m}/{\text{s}}^{\text{2}}$ for $a_h$, $3.00\text{m}/{\text{s}}^{\text{2}}$ for $a_v$ and $60.0°$ for $\varphi$ in equation (III).

  $\begin{array}{c}a=\sqrt{{\left(4.00\text{m}/{\text{s}}^{\text{2}}\right)}^2+{\left(3.00\text{m}/{\text{s}}^{\text{2}}\right)}^2+2\left(4.00\text{m}/{\text{s}}^{\text{2}}\right)\left(3.00\text{m}/{\text{s}}^{\text{2}}\right)\left(\cos 60.0°\right)}\\ =\sqrt{16.00+9.00+12.00}\text{m}/{\text{s}}^{\text{2}}\\ =6.08\text{m}/{\text{s}}^{\text{2}}\end{array}$

Substitute $4.00\text{m}/{\text{s}}^{\text{2}}$ for $a_h$, $60.0°$ for ${\varphi }_2$ and $3.00\text{m}/{\text{s}}^{\text{2}}$ for $a_v$ in equation (V).

  $\begin{array}{c}\tan {\theta }_2=\frac{\left(3.00\text{m}/{\text{s}}^{\text{2}}\right)\sin 60.0°}{\left(4.00\text{m}/{\text{s}}^{\text{2}}\right)+\left(3.00\text{m}/{\text{s}}^{\text{2}}\right)\cos 60.0°}\\ {\theta }_2={\tan }^{-1}\left(\frac{3\sqrt{3}}{11}\right)\\ =25.28°\\ \approx 25.3°\end{array}$

Thus, the acceleration of the object is $6.08\text{m}/{\text{s}}^{\text{2}}$ at an angle of $25.3°$ from horizontal.