Chapter 4, Problem 73P

(a)

To determine

The tension in the rope.

Answer to Problem 73P

The tension in the rope is $T=88\text{N}$

Explanation of Solution

Write the expression for the force for crate in terms of Newton’s second law along $x$ direction.

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

Here, $F_x$ is the frictional force, $f_k$ is the friction, $T$ is the tension, $m_1$ is the mass of the crate, $a_x$ is the acceleration along $x$ direction, and $g$ is the acceleration due to gravity.

Similarly write the expression for the force for the crate along $y$ direction.

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

Here, $F_y$ is the frictional force, $N$ is the normal force, $m_1$ is the mass of the crate, and $g$ is the acceleration due to gravity.

Write the expression for force for the box along  $x$ direction.

  ${\displaystyle \sum F_x=0}$        (III)

Similarly write the expression for the force for the box along $y$ direction.

  ${\displaystyle \sum F_y}=T-m_{2}g=m_2{a}_y$

Here, $F_y$ is the frictional force, $m_2$ is the mass of the box, $g$ is the acceleration due to gravity, $a_y$ is the acceleration along $y$ direction and, $T$ is the tension.

According to the problem $a_x=-a_y$ , and solve for $a_x$ from expression (I) and (II),

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

Solve for $a_y$ from the expression (IV)

  $\begin{array}{c}T-m_{2}g=m_2{a}_y\\ a_y=\frac{T}{m_2}-g\end{array}$        (VI)

Conclusion:

Eliminate for $a_x$ and $a_y$  and solve for $T$ , by equating the expression (V) and (VI),

  $\begin{array}{c}\frac{T}{m_1}+{\mu }_{k}g\cos \theta -g\sin \theta =\frac{T}{m_2}-g\\ T\left(\frac{1}{m_1}+\frac{1}{m_2}\right)=g\left(1+\sin \theta -{\mu }_k\cos \theta \right)\\ T=\frac{m_1{m}_2}{m_1+m_2}\left(1+\sin \theta -{\mu }_k\cos \theta \right)\end{array}$        (VII)

 Substitute $1\text{5kg}$ for $m_1$ , $8.0\text{kg}$ for $m_2$ , $\text{9}{\text{.8m/s}}^{\text{2}}$ for $g$ , $60°$ for $\theta$ , and $0.30$ for ${\mu }_k$ in expression (VII)

$\begin{array}{c}T=\frac{\left(1\text{5kg}\right)\left(8.0\text{kg}\right)}{\left(1\text{5kg}\right)+\left(8.0\text{kg}\right)}\left(1+\sin \left(60°\right)-\left(0.30\right)\cos \left(60°\right)\right)\\ =88\text{N}\end{array}$

The tension in the rope is $T=88\text{N}$

(b)

To determine

The time required by the crate to slide down.

Answer to Problem 73P

The time required by the crate to slide down is $2\text{s}$.

Explanation of Solution

Write the equation of motion, where $v_0=0$

  $\Delta x=-\frac{1}{2}{a}_x{t}^2$        (I)

Here, $\Delta x$ is the displacement, $a_x$ is the acceleration, and $t$ is the time

Re-write the expression in terms of $t$, and substitute $g-\frac{T}{m_2}$ from expression (VI) in sub part (a) as $a_x=-a_y$

  $\begin{array}{l}t=\sqrt{-\frac{2\Delta x}{a_x}}\\ t=\sqrt{-\frac{2\Delta x}{g-\frac{T}{m_2}}}\end{array}$        (II)

Conclusion:

Substitute $2.00\text{m}$ for $\Delta x$ , $88\text{N}$ for $T$, , $8.0\text{kg}$ for $m_2$ , and $\text{9}{\text{.8m/s}}^{\text{2}}$ for $g$ expression (II),

  $\begin{array}{c}t=\sqrt{-\frac{2\left(-2.00\text{m}\right)}{\text{9}{\text{.8m/s}}^{\text{2}}-\frac{88\text{N}}{8.0\text{kg}}}}\\ =2\text{s}\end{array}$

The time required by the crate to slide down is $2\text{s}$.

(c)

To determine

The force to push the crate.

Answer to Problem 73P

The force to push the crate is $\text{70N}$.

Explanation of Solution

Write the expression for the force for crate in terms of Newton’s second law along $x$ direction.

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

Here, $F_x$ is the frictional force, $P$ is the push, $f_k$ is the friction, $T$ is the tension, $m_1$ is the mass of the crate, $a_x$ is the acceleration along $x$ direction, and $g$ is the acceleration due to gravity.

Similarly write the expression for the force for the box along $y$ direction.

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

Here, $F_y$ is the frictional force, $m_2$ is the mass of the box, $T$ is the tension, and $g$ is the acceleration due to gravity.

Solve for $P$ by substituting for $T$ and $f_k$ from expression (I) and (II),

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

Conclusion:

Substitute $1\text{5kg}$ for $m_1$ , $8.0\text{kg}$ for $m_2$ , $\text{9}{\text{.8m/s}}^{\text{2}}$ for $g$ , $60°$ for $\theta$ , and $0.30$ for ${\mu }_k$ in expression (VII)

  $\begin{array}{c}P=\left(\text{9}{\text{.8m/s}}^{\text{2}}\right)\left(1\text{5kg}\left(0.30\cos \left(60°\right)+\sin \left(60°\right)\right)-8.0\text{kg}\right)\\ =70\text{N}\end{array}$

The force to push the crate is $\text{70N}$.

(d)

To determine

The smallest mass to keep the crate from sliding down.

Answer to Problem 73P

The smallest mass is $m_2=10\text{kg}$

Explanation of Solution

Write the expression for the force for crate in terms of Newton’s second law along $x$ direction.

  ${\displaystyle \sum F_x}=T+f_k-m_{1}g\sin \theta =0$        (I)

Here, $F_x$ is the frictional force, $f_k$ is the friction, $T$ is the tension, $m_1$ is the mass of the crate, $a_x$ is the acceleration along $x$ direction, and $g$ is the acceleration due to gravity.

Similarly write the expression for the force for the crate along $y$ direction.

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

Here, $F_y$ is the frictional force, $N$ is the normal force, $m_1$ is the mass of the crate, and $g$ is the acceleration due to gravity.

Write the expression for force for the box along  $x$ direction.

  ${\displaystyle \sum F_x=0}$        (III)

Similarly write the expression for the force for the box along $y$ direction.

  ${\displaystyle \sum F_y}=T-m_{2}g=0$        (IV)

Here, $F_y$ is the frictional force, $m_2$ is the mass of the box, $g$ is the acceleration due to gravity, $a_y$ is the acceleration along $y$ direction and, $T$ is the tension.

Conclusion:

Thus, $m_2=\frac{T}{g}$ , and find for $T$

  $\begin{array}{c}T=m_{1}g\sin \theta -fs\\ =m_{1}g\sin \theta -{\mu }_s{m}_{1}g\cos \theta \end{array}$        (V)

Thus, substitute the expression (V) in expression (IV) and re-write the expression in terms $m_2$ 

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

Substitute $1\text{5kg}$ for $m_1$ , $\text{9}{\text{.8m/s}}^{\text{2}}$ for $g$ , $60°$ for $\theta$ , and $0.40$ for ${\mu }_s$ in expression (IV)

  $\begin{array}{c}{m}_2=1\text{5kg}\left(\sin \left(60°\right)-0.40\cos \left(60°\right)\right)\\ =10\text{kg}\end{array}$

The smallest mass is $m_2=10\text{kg}$