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Q5.34(b):

Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig. 5.21). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are (b) the action-reaction forces between A and B ? What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between $\mu_s$ and $\mu_k$.

Solution :

Given Variables & System Configuration

  • Mass of body A ($m_A$) = $5\text{ kg}$
  • Mass of body B ($m_B$) = $10\text{ kg}$
  • Coefficient of friction ($\mu$) = $0.15$
  • Applied Force ($F$) = $200\text{ N}$
  • Acceleration due to gravity ($g$) = $9.8\text{ m/s}^2$

Step 1: Analysis of Static Equilibrium (Wall present)

When the system is resting against the wall, the forces are in equilibrium. The applied force $F$ is countered by the wall's reaction force and the frictional forces acting on bodies A and B.

Total Normal Force ($N_{total}$) = $(m_A + m_B)g = (5 + 10) \times 9.8 = 147\text{ N}$.

Total Frictional force ($f_{max}$) = $\mu N_{total} = 0.15 \times 147 = 22.05\text{ N}$.

Since the applied force $F = 200\text{ N}$ is greater than the total frictional resistance ($200 > 22.05$), the system would slide if not for the wall. The wall exerts a force $F_{wall}$ such that $F = F_{wall} + f_A + f_B$.

Step 2: Calculating Action-Reaction between A and B

The action-reaction force ($R$) is the force that body A exerts on B, and conversely, the force B exerts on A. Considering body B, the forces acting on it are the contact force $R$ (from A) and the frictional force $f_B$ (opposing motion). The wall exerts a reaction force on B.

For body B, the force applied by A is $R$. The frictional force on B is $f_B = \mu m_B g = 0.15 \times 10 \times 9.8 = 14.7\text{ N}$.

Since the bodies are in contact and pushed against a rigid wall, the net force on B is zero [Per Newton’s First Law]. Thus, the reaction force $R$ from A on B must overcome the friction on B: $R = f_B = 14.7\text{ N}$.

A (5kg) B (10kg) F=200N Wall

Step 3: Removing the Wall and Dynamic State

When the wall is removed, the system moves with acceleration $a$.

Net Force ($F_{net}$) = $F - f_{total} = 200 - 22.05 = 177.95\text{ N}$.

Acceleration ($a$) = $\frac{F_{net}}{m_A + m_B} = \frac{177.95}{15} \approx 11.86\text{ m/s}^2$.

To find the new force $R'$ between A and B, we consider the equation of motion for body B:

$R' - f_B = m_B a$

$R' = m_B a + f_B = (10 \times 11.86) + 14.7 = 118.6 + 14.7 = 133.3\text{ N}$.

Summary Analysis

Scenario Action-Reaction Force ($R$)
Wall Present (Static) $14.7\text{ N}$
Wall Removed (Dynamic) $133.3\text{ N}$

Final Solution: The action-reaction force is $14.7\text{ N}$ when the wall is present. When the wall is removed, the bodies move with an acceleration of $11.86\text{ m/s}^2$, and the action-reaction force changes to $133.3\text{ N}$.


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