Chapter 4, Problem 44P

To determine

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

Answer to Problem 44P

Solution:To prevent motion, the box A needs to have a mass of at least 6.7 kg.

Explanation of Solution

Given:

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

The coefficient of static friction, ${\mu }_s=0.30$

Formula:

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.

  

Figure.1

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

Draw the free body diagram of the system.

For box B

  

As per the condition applied, the box B neither moves up nor down,

If there is no motion;

  $m_{B}g=F_T$   ......(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),

  $\begin{array}{c}{F}_{fr}={\mu }_s{F}_N\\ ={\mu }_s{m}_{A}g\end{array}$   ......(4)

From equation (1) and (3),

  $F_{fr}=m_{B}g$

From equation (4),

  $m_{B}g={\mu }_s{m}_{A}g$

Simplify for $m_A$ .

  $m_A=\frac{m_B}{{\mu }_s}$

Calculation:

Substitute the given values in the formula to calculate $m_A$ .

  $\begin{array}{c}{m}_A=\frac{m_B}{{\mu }_s}\\ =\frac{2.0\text{ kg}}{0.30}\\ =6.666\text{ kg}\end{array}$

Conclusion:

If there is to be no motion associated with the boxes, then the mass of the box A must at least be 6.7 kg.