Chapter 4, Problem 122P

(a)

To determine

The acceleration of each blocks and tension in the cord.

Answer to Problem 122P

The magnitude of acceleration of each blocks and tension in the cord is respectively $\left(\frac{m_2-{\mu }_k{m}_1}{m_1+m_2}\right)g$ and $\left({\mu }_k+1\right)\frac{m_1{m}_2}{m_1+m_2}g$.

Explanation of Solution

Case 1 (mass $m_1$).

Write the expression for the equilibrium condition in y-direction (Vertical).

$N=m_{1}g$ (I)

Here, N is the normal force on the mass $m_1$ and g is the acceleration due to gravity.

Write the expression for the equilibrium condition in x-direction (Horizontal).

$T-{\mu }_k{m}_{1}g=m_1{a}_x$ . (II)

Here, $a_x$ is the acceleration of mass $m_1$ in the horizontal direction and T is the tension in the cord.

Rewrite the above equation in terms of $a_x$.

$a_x=\frac{T}{m_1}-{\mu }_{k}g$ . (III)

Case 2 (mass $m_2$).

Write the expression for the equilibrium condition in y-direction (Vertical).

$T-m_{2}g=m_2{a}_y$ (IV)

Here, $a_y$ is the acceleration of mass $m_2$.

Write the above expression in terms of $a_y$.

$a_y=\frac{T}{m_2}-g$ . (V)

Since both acceleration is equal in magnitude and opposite in direction.

Write the expression for the relation between acceleration of two blocks.

$a_x=-a_y$ (VI)

Substitute (III) and (V) in the above equation to calculate T.

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

Substitute (VI) in (v) to calculate the acceleration $a_y$.

$\begin{array}{c}{a}_y=\frac{1}{m_2}\left(\left({\mu }_k+1\right)\frac{m_1{m}_2}{m_1+m_2}g\right)-g\\ =\left({\mu }_k+1\right)\frac{m_1}{m_1+m_2}g-g\\ =g\left(\frac{{\mu }_k{m}_1+m_1}{m_1+m_2}-1\right)\\ =\frac{{\mu }_k{m}_1-m_2}{m_1+m_2}g\end{array}$

Substitute the above expression in (VI) to calculate $a_x$.

$\begin{array}{l}{a}_x=-\left(\frac{{\mu }_k{m}_1-m_2}{m_1+m_2}g\right)\\ a_x=\left(\frac{m_2-{\mu }_k{m}_1}{m_1+m_2}\right)g\end{array}$ . (VIII)

Conclusion:

Therefore, the magnitude of acceleration of each blocks and tension in the cord is respectively $\left(\frac{m_2-{\mu }_k{m}_1}{m_1+m_2}\right)g$ and $\left({\mu }_k+1\right)\frac{m_1{m}_2}{m_1+m_2}g$.

(b)

To determine

The magnitude of acceleration and tension for the cases $m_1<<m_2$, $m_1>>m_2$ and $m_1=m_2$.

Answer to Problem 122P

The magnitude of acceleration and tension for the cases $m_1<<m_2$, $m_1>>m_2$ and $m_1=m_2$ is respectively is zero or free fall and $g$, $m_{2}g$ and $0$, $\left({\mu }_k+1\right)\frac{m_{2}g}{2}$ and $\left(\frac{\left(1-{\mu }_k\right)}{2}\right)g$.

Explanation of Solution

Consider the equation (VII) and (VIII).

Case 1 $m_1<<m_2$.

$\begin{array}{c}T=\left({\mu }_k+1\right)\frac{m_1{m}_2}{m_1+m_2}g\\ =\left({\mu }_k+1\right)\frac{m_{1}g}{\left(1+\frac{m_1}{m_2}\right)}\\ \sim \left({\mu }_k+1\right)m_{1}g\end{array}$

Since ${\mu }_k$ is too small, ${\mu }_k$ could possible to neglect.

Thus, the tension in the cord is $m_{1}g$ and is negligible might be considered as free-fall.

$\begin{array}{c}{a}_x=\left(\frac{m_2-{\mu }_k{m}_1}{m_1+m_2}\right)g\\ =\left(\frac{\frac{m_2}{m_1}-{\mu }_k}{\left(1+\frac{m_2}{m_1}\right)}\right)g\end{array}$

Since $1+\frac{m_2}{m_1}$ is $\frac{m_2}{m_1}$ and ${\mu }_k$ is too small, ${\mu }_k$ could possible to neglect.

Thus, the acceleration of the block is $g$.

Case 2 $m_1>>m_2$.

$\begin{array}{c}T=\left({\mu }_k+1\right)\frac{m_1{m}_2}{m_1+m_2}g\\ =\left({\mu }_k+1\right)\frac{m_{2}g}{\left(1+\frac{m_2}{m_1}\right)}\\ \sim \left({\mu }_k+1\right)m_{2}g\end{array}$

Since ${\mu }_k$ is too small, ${\mu }_k$ could possible to neglect.

Thus, the tension in the cord is $m_{2}g$.

$\begin{array}{c}{a}_x=\left(\frac{m_2-{\mu }_k{m}_1}{m_1+m_2}\right)g\\ =\left(\frac{\frac{m_2}{m_1}-{\mu }_k}{\left(1+\frac{m_2}{m_1}\right)}\right)g\end{array}$

Since $1+\frac{m_2}{m_1}$ is $\frac{m_2}{m_1}$ and ${\mu }_k$ is too small, ${\mu }_k$ could possible to neglect.

Thus, the acceleration of the block is $0$.

Case 3 $m_1=m_2$.

Substitute $m_1=m_2$ in (VII) and (VIII) to calculate T and $a_x$.

$\begin{array}{c}T=\left({\mu }_k+1\right)\frac{m_2{m}_2}{m_2+m_2}g\\ =\left({\mu }_k+1\right)\frac{m_2{}^{2}g}{\left(2m_2\right)}\\ =\left({\mu }_k+1\right)\frac{m_{2}g}{2}\end{array}$

$\begin{array}{c}{a}_x=\left(\frac{m_2-{\mu }_k{m}_2}{m_2+m_2}\right)g\\ =\left(\frac{m_2\left(1-{\mu }_k\right)}{2m_2}\right)g\\ =\left(\frac{\left(1-{\mu }_k\right)}{2}\right)g\end{array}$

Conclusion:

Therefore, the magnitude of acceleration and tension for the cases $m_1<<m_2$, $m_1>>m_2$ and $m_1=m_2$ is respectively is zero or free fall and $g$, $m_{2}g$ and $0$, $\left({\mu }_k+1\right)\frac{m_{2}g}{2}$ and $\left(\frac{\left(1-{\mu }_k\right)}{2}\right)g$.

(c)

To determine

The magnitude of $m_2$ and tension in the cord.

Answer to Problem 122P

The magnitude of $m_2$ and tension in the cord is respectively ${\mu }_k{m}_1$ and $m_{2}g$.

Explanation of Solution

Consider the equation (VII) and (VIII).

In the case of constant velocity acceleration should be zero.

Equate (VIII) to zero to calculate $m_2$.

$\begin{array}{c}\left(\frac{m_2-{\mu }_k{m}_1}{m_1+m_2}\right)g=0\\ m_2-{\mu }_k{m}_1=0\\ m_2={\mu }_k{m}_1\end{array}$

Substitute the above equation for $m_2$ in (VII) to calculate T.

$\begin{array}{c}T=\left({\mu }_k+1\right)\frac{m_1{m}_2}{m_1+\left({\mu }_k{m}_1\right)}g\\ =\left({\mu }_k+1\right)\frac{m_1{m}_2}{m_1\left(1+{\mu }_k\right)}g\\ =m_{2}g\end{array}$

Conclusion:

Therefore, the magnitude of $m_2$ and tension in the cord is respectively ${\mu }_k{m}_1$ and $m_{2}g$.