Animated Solution for Physics - Laws of Motion: Given in the figure are two blocks $A$ and $B$ of weight $20\text{ N}$ and $100\text{ N}$ respectively. These are being pressed against a wall by a force $F$ as shown in figure. If the coefficient of friction between the blocks is $0.1$ and between block $B$ and the wall is $0.15$, the frictional force applied by the wall in block $B$ is

Imagine you are holding two heavy books against a wall by pressing them with your hand. This is exactly what is happening in our problem. We have two blocks, $A$ and $B$, with weights $20\text{ N}$ and $100\text{ N}$ respectively, being pressed against a wall by a horizontal force $F$.

The problem generously provides the coefficients of friction: $0.1$ between the blocks and $0.15$ between block $B$ and the wall. But beware! This is a classic JEE trap. Since the blocks are in static equilibrium (they are not sliding), the static friction is a self-adjusting force. It will only grow as large as it needs to be to prevent motion. The formula $f_{max}=\mu N$ only tells us the maximum possible friction, not the actual friction acting right now. As long as the force $F$ is strong enough, we can completely ignore the coefficients of friction and rely purely on balancing the forces.

To unravel this, we must look at the system piece by piece. Let's draw a Free Body Diagram (FBD) for Block $A$.

In the vertical direction, gravity is pulling Block $A$ downwards with a force equal to its weight, $W_A=20\text{ N}$. Since Block $A$ is not falling, there must be an upward force perfectly balancing this weight. This savior is the static friction force exerted by Block $B$ on Block $A$. Let's call it $f_A$.

By applying the equilibrium condition $\sum F_y=0$, we get:

So, Block $B$ is holding Block $A$ up with a force of $20\text{ N}$.

Now, let's shift our focus to Block $B$. This is where Newton's Third Law of Motion comes into play. Every action has an equal and opposite reaction. If Block $B$ exerts an upward frictional force of $20\text{ N}$ on Block $A$, then Block $A$ must exert a downward frictional force of $20\text{ N}$ on Block $B$.

So, what are the downward forces acting on Block $B$? 1. Its own weight, $W_B=100\text{ N}$. 2. The downward friction from Block $A$, $f_A=20\text{ N}$.

Total downward force on Block $B$ is $100\text{ N}+20\text{ N}=120\text{ N}$.

Block $B$ is also in equilibrium; it isn't sliding down the wall. Therefore, the wall must be exerting an upward frictional force, let's call it $f_B$, to perfectly balance all the downward forces acting on Block $B$.

Applying the equilibrium condition for Block $B$:

Substituting the values we found:

This elegant solution reminds us to always trust the fundamental laws of equilibrium before blindly plugging numbers into formulas. The extra data was just an illusion!