Forces Acting on a Ladder Leaning Against Walls

The question asks us to identify the direction of the force of friction acting on a ladder that is placed on a smooth ground and leaning against a rough vertical wall.

Identifying the Forces

To solve this, we need to consider all the forces acting on the ladder when it's in a stable position (equilibrium):

• Weight ($W$): This force acts vertically downwards, typically assumed to be at the center of the ladder. We can represent it as $W = mg$, where $m$ is the mass of the ladder and $g$ is the acceleration due to gravity.
• Normal Force from Ground ($N_g$): The ground pushes upwards on the ladder's lower end, perpendicular to the ground surface.
• Friction Force from Ground ($f_g$): This force would act horizontally at the lower end, opposing any tendency to slip. However, the problem states the ground is smooth, which means $f_g = 0$. The ground cannot provide any frictional force.
• Normal Force from Wall ($N_w$): The wall pushes horizontally outwards on the ladder's upper end, perpendicular to the wall surface.
• Friction Force from Wall ($f_w$): This force acts vertically at the upper end, where the ladder contacts the wall. Since the wall is stated to be rough, it can exert a frictional force.

Applying Equilibrium Conditions

For the ladder to remain in place without moving or rotating, the net force and net torque acting on it must be zero.

Let's consider the torques about the point where the ladder touches the ground (the lower end). Let $\theta$ be the angle the ladder makes with the ground.

• The weight ($W$) acts downwards at the ladder's midpoint. The horizontal distance from the ground contact point to the point where weight acts is $(\frac{L}{2})\cos\theta$, where $L$ is the length of the ladder. This creates a clockwise torque: $\tau_W = W \times (\frac{L}{2})\cos\theta$.
• The normal force from the wall ($N_w$) acts horizontally outwards at the top end. The vertical height of the top end from the ground is $L\sin\theta$. This force also creates a clockwise torque about the ground contact point: $\tau_{Nw} = N_w \times (L\sin\theta)$.
• The friction force from the wall ($f_w$) acts vertically at the top end. If it acts upwards, its horizontal distance (lever arm) from the ground contact point is $L\cos\theta$. This creates a counter-clockwise torque: $\tau_{fw} = f_w \times (L\cos\theta)$.

For the ladder to be in rotational equilibrium, the counter-clockwise torque must balance the clockwise torques:

We can simplify this by dividing by $L$:

Now, let's solve for $f_w$:

From this equation, we see that $W$ (weight) is positive, $N_w$ (normal force from the wall) is positive, and $\tan\theta$ is positive for typical angles of a leaning ladder ($0 < \theta < 90^\circ$). Therefore, $f_w$ must be a positive value.

Determining the Direction of Friction

A positive value for $f_w$ in our torque equation means the friction force must act in the direction we assumed to create the counter-clockwise torque. We assumed friction acted upwards at the upper end. This upward friction counteracts the tendency of the ladder to slip downwards at the top due to its weight and the outward push from the wall.

Key Point: Even though the ground is smooth and cannot provide friction, the rough wall must provide friction to maintain balance. The ladder tends to slide down the wall, so friction opposes this by acting upwards.

Conclusion

Based on the torque analysis, the friction force acting on the ladder at the point of contact with the rough vertical wall must be directed upwards.