Chapter 4, Problem 45P

To determine

(a) To Determine:

The minimum mass of the box A to prevent any motion from occurring.

Answer to Problem 45P

Solution:

To prevent motion, the box A needs to have a mass of at least 5.0 kg.

Explanation of Solution

Boxes A and B are connected by a rope passing over a pulley. The box A lies on a table and the box B hangs freely. The box A has its weight $m_{A}g$ acting vertically downwards and the normal force $F_N$ acts vertically upwards due to Newton’s third law. Since the box is connected to the box B, it also experiences the tension force $F_T$ along the length of the rope.

The box B hangs freely. Its weight $m_{B}g$ acts vertically downwards and since, the same rope passes over the pulley connecting the boxes A and B, the same tension force $F_T$ acts on the box B along the length of the rope. The forces act as shown in the diagram given below.

The boxes will not move if the force of static friction acting between the box A and the table is at least equal to the tension acting on the box.

The boxes will move with constant speed when the force of kinetic friction acting between the box A and the table is equal to the tension acting on the box.

Draw the free body diagram of the system.

For box B

Since the condition applied is that there is no motion, the box B neither moves up nor down.

Therefore,

    $m_{B}g=F_T$…………(1)

Draw the free body diagram for the box A.

For vertical equilibrium,

    $m_{A}g=F_N$…………(2)

For horizontal equilibrium,

    $F_{fr}=F_T$…………(3)

The force of static friction and the normal force are related by the expression,

    $F_{fr}={\mu }_s{F}_N$

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

From equation (2)

    $\begin{array}{c}{F}_{fr}={\mu }_s{F}_N\end{array}$

          $\begin{array}{c}={\mu }_s{m}_{A}g\end{array}$…………(4)

From equation (1)and (3),

    $F_{fr}=m_{B}g$

From equation (4),

    $m_{B}g={\mu }_s{m}_{A}g$

Simplify for $m_A$.

    $m_A=\frac{m_B}{{\mu }_s}$

Given:

The mass of the boxB$m_B=2.0\text{ kg}$

The coefficient of static friction ${\mu }_s=0.40$

The coefficient of kinetic friction ${\mu }_k=0.20$

Formula:

    $m_A=\frac{m_B}{{\mu }_s}$

Calculation:

Substitute the given values in the formula (a)to calculate $m_A$

when the boxes are at rest.

    $\begin{array}{c}{m}_A=\frac{m_B}{{\mu }_s}\end{array}$

          $\begin{array}{c}=\frac{2.0\text{ kg}}{0.40}\end{array}$

          $\begin{array}{c}=5.0\text{ kg}\end{array}$

To determine

(b) To Determine:

The value of the mass of the box A so that the system moves with constant velocity.

Answer to Problem 45P

Solution:

The mass of the box A so that the system moves with a constant speed is $1.0\times {10}^1\text{ kg}$

Explanation of Solution

If the boxes move with constant speed, then the resultant force on the boxes is also zero. Therefore, the minimum mass of the box Ais obtained by replacing the coefficient of static friction in the above expression by the coefficient of kinetic friction ${\mu }_k$.

For the boxes to move with constant speed,

    $m_A=\frac{m_B}{{\mu }_k}$

Given:

The mass of the box B$m_B=2.0\text{ kg}$

The coefficient of static friction ${\mu }_s=0.40$

The coefficient of kinetic friction ${\mu }_k=0.20$

Formula used:

    $m_A=\frac{m_B}{{\mu }_k}$

Calculation:

Substitute the given values in the formula (b)to calculate $m_A$ when the boxes move with constant speed.

    $\begin{array}{c}{m}_A=\frac{m_B}{{\mu }_k}\end{array}$

            $\begin{array}{c}=\frac{2.0\text{ kg}}{0.20}\end{array}$

            $\begin{array}{c}=1.0\times {10}^1\text{ kg}\end{array}$