Chapter 4, Problem 175P

(a)

To determine

Tension in the rope.

Answer to Problem 175P

The tension is $\underset{\_}{88\text{N}}$.

Explanation of Solution

Using Newton’s second law, write the equation for force on crate on x direction.

${\displaystyle \sum F_x=T+f_k-m_{1}g\sin \theta }=m_1{a}_x$        (I)

Here $T$ is the tension, $f_k$ is the force due to kinetic friction, $m_1$ is the mass of the crate, $g$ is the acceleration due to gravity, $a_x$ is the acceleration and $\theta$ is the inclination angle

Write the equation for force in the y-direction

${\displaystyle \sum F_y=N-m_{1}g\cos \theta }=0$        (II)

Here $N$ is the normal force on the crate.

Write the equation for force on the box in x and y direction

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ {\displaystyle \sum F_y=T-m_2{a}_y}\end{array}$        (III)

Here $m_2$ is the mass of the box

Solve for $a_x$ by assuming $a_x=-a_y$

  $\begin{array}{c}{m}_1{a}_x=T+{\mu }_{k}N-m_{1}g\sin \theta \\ =T+{\mu }_k{m}_{1}g\cos \theta -m_{1}g\sin \theta \\ a_x=\left(\frac{T}{m_1}\right)+{\mu }_{k}g\cos \theta -g\sin \theta =-a_y\end{array}$

Here ${\mu }_k$ is the coefficient of kinetic friction.

Solve for $a_y$

  $\begin{array}{c}{m}_2{a}_y=T-m_{2}g\\ a_y=\frac{T}{m_2}-g\end{array}$

Eliminate $a_x$ and $a_y$ and solve for $T$

  $\begin{array}{c}g-T/m_2=T/m_1+{\mu }_{k}g\cos \theta -g\sin \theta \\ T=\frac{m_1{m}_{2}g}{m_1+m_2}\left(1+\sin \theta -{\mu }_k\cos \theta \right)\end{array}$        (III)

Conclusion:

Substitute $15\text{kg}$ for $m_1$, $8\text{kg}$ for $m_2$, $9.\text{8}\text{m}/{\text{s}}^{\text{2}}$ for $g$, $60°$ for $\theta$ and $0.30$ for ${\mu }_k$ in (III)

$\begin{array}{c}T=\frac{\left(15\text{kg}\right)\left(8\text{kg}\right)\left(9.\text{8}\text{m}/{\text{s}}^{\text{2}}\right)}{15\text{kg}+8\text{kg}}\left(1+\sin \left(60°\right)-0.30\cos 60°\right)\\ =88\text{N}\end{array}$

The tension is $\underset{\_}{88\text{N}}$.

(b)

To determine

Time took by the crate to slide down the incline.

Answer to Problem 175P

The time is $\underset{\_}{2\text{s}}$

Explanation of Solution

The crate begins from rest. Then initial velocity is zero.

Refer subpart a and write the equation for acceleration

$\begin{array}{l}{a}_x=-a_y\\ =g-\frac{T}{m_2}\end{array}$        (IV)

Write the equation of motion for distance covered

$\Delta x=\frac{1}{2}{a}_x{\left(\Delta t\right)}^2$        (V)

Here $\Delta t$ is the time

Substitute (IV) in (V) and rewrite in terms of $\left(\Delta t\right)$

$\Delta t=\sqrt{\frac{2\Delta x}{g-\frac{T}{m_2}}}$        (VI)

Conclusion

Substitute $8\text{kg}$ for $m_2$, $9.\text{8}\text{m}/{\text{s}}^{\text{2}}$ for $g$, $88\text{N}$ for $T$ and $2\text{m}$ for $\Delta x$ in (VI)

$\begin{array}{c}\Delta t=\sqrt{\frac{2\left(-2\text{m}\right)}{9.\text{8}\text{m}/{\text{s}}^{\text{2}}-\frac{88\text{N}}{8\text{kg}}}}\\ =2\text{s}\end{array}$

The time is $\underset{\_}{2\text{s}}$

(c)

To determine

Force to push the crate.

Answer to Problem 175P

The force is $\underset{\_}{70\text{N}}$.

Explanation of Solution

Let $P$ be the force with which you push on the crate.

Then,

$\begin{array}{c}{\displaystyle \sum F_x=T-P-f_k-m_{1}g\sin \theta }=0\\ P=f_k+m_{1}g\sin \theta -T\end{array}$        (VII)

Write the force on the box

$\begin{array}{c}{\displaystyle \sum F_y=T-m_{2}g=0}\\ T=m_{2}g\end{array}$        (VIII)

Substitute for $T$ and $f_k$ in (VII)

$\begin{array}{c}P={\mu }_k{m}_{1}g\cos \theta +m_{1}g\sin \theta -m_{2}g\\ =g\left(m_1\left({\mu }_k\cos \theta +\sin \theta \right)-m_2\right)\end{array}$        (IX)

Conclusion

Substitute $15\text{kg}$ for $m_1$, $8\text{kg}$ for $m_2$, $9.\text{8}\text{m}/{\text{s}}^{\text{2}}$ for $g$, $60°$ for $\theta$ and $0.30$ for ${\mu }_k$ in (IX)

$\begin{array}{c}P=9.\text{8}\text{m}/{\text{s}}^{\text{2}}\left(15\text{kg}\left(0.30\cos 60°+\sin 60°\right)-8\text{kg}\right)\\ =70\text{N}\end{array}$

The force is $\underset{\_}{70\text{N}}$.

(d)

To determine

Smallest mass that you could substitute to keep the crate from sliding down the incline.

Answer to Problem 175P

The mass is $\underset{\_}{10\text{kg}}$.

Explanation of Solution

The net force on each object would be zero.

Then net force for the crate,

$\begin{array}{l}{\displaystyle \sum F_x=T+{\mu }_s{m}_{1}g\cos \theta -m_{1}g\sin \theta =0}\\ {\displaystyle \sum F_y=N-m_{1}g\cos \theta =0}\end{array}$        (X)

Here ${\mu }_s$ is the coefficient of static friction.

Force of the box,

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ {\displaystyle \sum F_y=T-m_{2}g=0}\end{array}$        (XI)

Then,

$m_2=\frac{T}{g}$        (XII)

Tension is,

$T=m_{1}g\sin \theta -{\mu }_s{m}_{1}g\cos \theta$        (XIII)

Substitute for $T$ from (XIII) in (XII)

$\begin{array}{c}{m}_2=\frac{m_{1}g\sin \theta -{\mu }_s{m}_{1}g\cos \theta }{g}\\ =m_1\sin \theta -{\mu }_s{m}_1\cos \theta \end{array}$

Conclusion

Substitute $15\text{kg}$ for $m_1$, $60°$ for $\theta$ and $0.40$ for ${\mu }_s$ in (IX)

    $\begin{array}{c}{m}_2=15\text{kg}\sin 60°-\left(0.40\right)\left(15\text{kg}\right)\cos 60°\\ =10\text{kg}\end{array}$

The mass is $\underset{\_}{10\text{kg}}$.