Chapter 4, Problem 104P

(a)

To determine

The acceleration of the crate.

Answer to Problem 104P

The acceleration of the crate is $\underset{\_}{0.31\text{m}/{\text{s}}^2}$.

Explanation of Solution

Let $25.0\text{kg}$ be $m_1$, and $14.0\text{kg}$ be $m_2$.

Write the expression for the frictional force acting on mass $m_2$.

  $f_2=\mu m_{2}g$        (I)

Here, $\mu$ is the coefficient of friction, $g$ is the acceleration due to gravity.

Write the expression for frictional force acting on mass $m_1$.

  $f_1=\mu \left(m_{1}g-F\sin \theta \right)$        (II)

Here, $F$ is the applied force, $\theta$ is the angle inclined with mass $m_1$.

Write the expression for the total force acting on $m_2$.

  $T-f_2=m_{2}a$        (III)

Here, $T$ is the tension.

Substitute, equation (I) in (III), and rearrange to obtain an expression for $T$.

  $\begin{array}{c}T-\mu m_2{g}_2=m_{2}a\\ T=m_2\left(\mu g+a\right)\end{array}$        (IV)

Write the expression for total force acting on mass $m_1$.

  $F\cos \theta -T-\mu \left(m_{1}g-F\sin \theta \right)=m_{1}a$        (V)

Substitute equation (IV) in (V), and rearrange to obtain an expression for $a$.

  $\begin{array}{c}F\cos \theta -m_2\left(\mu g+a\right)-\mu \left(m_{1}g-F\sin \theta \right)=m_{1}a\\ a=\frac{F\left(\cos \theta -\mu \sin \theta \right)-\mu g\left(m_1+m_2\right)}{m_1+m_2}\end{array}$        (VI)

Conclusion:

Substitute, $195\text{N}$ for $F$, $20°$ for $\theta$, $0.55$ for $\mu$, $25.0\text{kg}$ for $m_1$, and $14.0\text{kg}$ for $m_2$, and $9.8\text{m}/{\text{s}}^2$ for $g$ in equation (VI).

  $\begin{array}{c}a=\frac{195\text{N}\left(\cos 20°+\mu \sin 20°\right)-\left(0.55\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(25.0\text{kg}+14.0\text{kg}\right)}{\left(25.0\text{kg}+14.0\text{kg}\right)}\\ =0.31\text{m}/{\text{s}}^2\end{array}$

Therefore, the acceleration of the crate is $\underset{\_}{0.31\text{m}/{\text{s}}^2}$.

(b)

To determine

The tension of the rope connecting two wires.

Answer to Problem 104P

The tension of the rope connecting two wires is $\underset{\_}{80\text{N}}$.

Explanation of Solution

Use equation (IV) to find the tension.

Conclusion:

Substitute, $14.0\text{kg}$ for $m_2$, $0.55$ for $\mu$, $0.31\text{m}/{\text{s}}^2$ for $a$, and $9.8\text{m}/{\text{s}}^2$ for $g$ in equation (IV) to find the tension.

  $\begin{array}{c}T=14.0\text{kg}\left(0.55\left(9.8\text{m}/{\text{s}}^2\right)+0.31\text{m}/{\text{s}}^2\right)\\ =80\text{N}\end{array}$

Therefore, the tension of the rope connecting two wires is $\underset{\_}{80\text{N}}$.

(c)

To determine

The distance moved by the crates in $3.0\text{s}$.

Answer to Problem 104P

The distance moved by the crates in $3.0\text{s}$ is $\underset{\_}{1.4\text{m}}$.

Explanation of Solution

Write the kinematic equation for distance moved.

  $s={\upsilon }_{0}t+\frac{1}{2}a{t}^2$        (VII)

Here, ${\upsilon }_0$ is the initial velocity.

Conclusion:

Substitute, $0$ for ${\upsilon }_0$, $0.31\text{m}/{\text{s}}^2$ for $a$ , and $3.0\text{s}$ for $t$ in equation (VII).

  $\begin{array}{c}s=0+\frac{1}{2}\left(0.31\text{m}/{\text{s}}^2\right){\left(3.0\text{s}\right)}^2\\ =1.4\text{m}\end{array}$

Therefore, the distance moved by the crates in $3.0\text{s}$ is $\underset{\_}{1.4\text{m}}$.