Animated Solution for Physics - Rotational Motion: A block of mass $m$ is at rest under the action of a force $F$ against a wall as shown in figure. Which of the following statement is incorrect? (2005)

Have you ever leaned against a wall and wondered about the invisible forces keeping you perfectly still? This problem presents a deceptively simple scenario: a block pressed against a wall by a horizontal force $F$. At first glance, it looks like a basic exercise in balancing forces. But hidden within this static setup is a beautiful dance of torques that catches many students off guard.

Translational Equilibrium

Let's start with what we can easily see. The block is at rest, which means it is in a state of translational equilibrium. According to Newton's Second Law, the net force acting on the block must be zero in all directions.

If we look at the vertical axis, gravity is relentlessly pulling the block downwards with a force of $mg$. To prevent the block from sliding down, the wall exerts an upward static frictional force $f$. Since these are the only two vertical forces, they must perfectly balance each other. Therefore, we can confidently state that $f=mg$.

Now, let's examine the horizontal axis. We are pushing the block into the wall with an applied force $F$. In response, the rigid wall pushes back with a normal reaction force $N$. Again, because the block isn't accelerating horizontally, these forces must be equal in magnitude. Thus, $F=N$.

Up to this point, everything feels intuitive. Statements (a) and (b) are mathematically sound. But physics is rarely just about pushing and pulling; it's also about twisting and turning.

Enter the Torque

Here is where the problem elevates from a standard textbook exercise to a true test of conceptual depth. For a rigid body to be completely at rest, it must also be in rotational equilibrium. This means the net torque (the turning effect of forces) about any point must be zero. Let's choose the center of mass $C$ as our pivot point.

Let's evaluate the torques produced by our four forces. The applied force $F$ is directed straight through the center of mass. Similarly, the gravitational force $mg$ acts exactly at the center of mass. Because their lines of action pass directly through our chosen pivot, their perpendicular distances are zero. Consequently, they produce absolutely zero torque (${\tau }_F=0$ and ${\tau }_{mg}=0$). Statement (c) survives the test.

But what about friction? The frictional force $f$ acts along the surface of the wall, which is at a distinct perpendicular distance from the center of mass. Because it points upwards along the left edge of the block, it creates a clockwise torque (${\tau }_f$) about the center.

If this were the end of the story, the block would start spinning clockwise! Since it doesn't, there must be a counteracting torque.

The Shifting Normal

This brings us to the grand finale and the downfall of statement (d). How does the block prevent itself from spinning? The answer lies in the behavior of the normal force $N$.

While we often draw the normal force acting straight through the center of an object, it is actually a distributed force that can shift its effective point of application. To counteract the clockwise torque generated by friction, the normal force $N$ dynamically shifts downwards, below the center of mass.

By acting below the center, the normal force creates an anticlockwise torque (${\tau }_N$). For the block to remain perfectly still, this anticlockwise torque must exactly cancel the clockwise torque from friction (${\tau }_N+{\tau }_f=0$).

Therefore, the normal force absolutely must produce a torque. Statement (d) claims that $N$ will not produce torque, making it the mathematically incorrect statement we were hunting for.

This problem is a masterful reminder: in the world of rigid body dynamics, never assume a force acts through the center of mass until you've checked the torques!