Question

A light ladder is supported on a rough floor and leans against a smooth wall, touching the wall at height \( h \) above the floor. A man climbs up the ladder until the base of the ladder is on the verge of slipping. The coefficient of static friction between the foot of the ladder and the floor is \( \mu \). The horizontal distance moved by the man is:

• A. \( \mu^2 h \)
• B. \( \frac{\mu}{h} \)
• C. \( \mu h \)
• D. \( \mu^2 h^2 \)

Hint:

The distance moved by a person on a ladder depends on the frictional force and the height at which the ladder touches the wall.

Answer 1

The Correct Option is C

To solve this problem, let's consider the forces and torques acting on the ladder. The ladder is supported on a rough floor and leans against a smooth wall. The key forces acting on the ladder are:

• The gravitational force acting at the center of the ladder. 
• The normal force \( N \) from the wall, which acts horizontally.
• The normal force \( N_f \) from the floor, which acts vertically.
• The frictional force \( f \) at the base of the ladder, acting horizontally.

The ladder is at the verge of slipping when the static frictional force \( f = \mu N_f \). Let \( L \) be the length of the ladder. The torque balance about the point where the ladder touches the floor gives us:
\[\begin{align*} N \cdot h + \mu N_f \cdot \frac{L}{2} = N_f \cdot \frac{L}{2} \end{align*}\]For the ladder to be in equilibrium, the sum of vertical forces and horizontal forces must be zero. Thus:
\[\begin{align*} N_f = W + Mg,\quad N = f = \mu N_f \end{align*}\]Since \( f = N \):
\[\begin{align*} \mu N_f = N \end{align*}\]Thus, \( N_f = \frac{N}{\mu} \). Substituting this into the torque balance equation results in:
\[\begin{align*} N \cdot h + \mu \cdot \frac{N}{\mu} \cdot \frac{L}{2} = \frac{N}{\mu} \cdot \frac{L}{2} \end{align*}\]Simplifying, this equation becomes:
\[\begin{align*} N \cdot h + \frac{N \cdot L}{2} = \frac{N \cdot L}{2 \mu} \end{align*}\]Rearranging and solving for \( L \) yields:
\[\begin{align*} N \cdot h = \frac{N \cdot L}{2 }(1 - \frac{1}{\mu}) \end{align*}\]Simplifying further, and recognizing that the remaining torque relation allows us to solve for \( L \):
\[\begin{align*} L = 2h \mu \end{align*}\]Therefore, the horizontal distance moved by the man:\[\mu \cdot h\]

Answer 2

In this case, the condition for slipping at the base of the ladder is when the frictional force equals the horizontal force. The torque about the base of the ladder will balance at the verge of slipping. The equation for the static friction force is: \[ f_{\text{friction}} = \mu \cdot N \] where \( N \) is the normal force. The horizontal force and the weight of the man on the ladder create a torque that balances out, and the horizontal distance moved by the man is given by: \[ d = \mu h \] Thus, the horizontal distance moved by the man is \( \mu h \).

Answer 3

Here we look at the equilibrium of the light ladder by taking torques about the point where the ladder touches the wall, rather than about the foot of the ladder. Let the wall be a vertical line and the floor be horizontal. The base of the ladder sits at a horizontal distance \( b \) from the wall, and the top touches the wall at height \( h \).

Since the ladder itself is massless, the floor must support the man's entire weight: the vertical reaction at the floor is \( N_f = mg \). At the verge of slipping, the friction force at the base is \( f = \mu N_f = \mu mg \), directed horizontally toward the wall. The wall, being smooth, pushes back horizontally with a reaction \( N_w \). Horizontal equilibrium gives \( N_w = f = \mu mg \).

Taking torques about the top contact point (where the ladder touches the wall), the wall's own reaction passes through this point and contributes no torque. The floor's normal reaction, acting vertically at the base, has a moment arm equal to the horizontal distance \( b \) from the wall, contributing a torque \( N_f \cdot b = mgb \). The friction force, acting horizontally at the base, has a moment arm equal to the height \( h \), contributing a torque \( f \cdot h = \mu mgh \) in the opposite rotational sense. The man's weight, acting vertically at his position (a horizontal distance \( b-d \) from the wall, where \( d \) is how far he has moved from the base), contributes a torque \( mg(b-d) \), also opposing the floor's normal-force torque.

Balancing torques about the top point:

\[ mgb = \mu mgh + mg(b-d) \]

Dividing through by \( mg \) and simplifying:

\[ b = \mu h + b - d \implies d = \mu h \]

Now checking each option:

1. Option A, \( \mu^2 h \): this would require an extra factor of \( \mu \) that does not appear anywhere in the torque balance above — the friction force enters linearly, not squared.
2. Option B, \( \frac{\mu}{h} \): this has the wrong dependence on \( h \) altogether (inversely rather than directly proportional) and is not dimensionally sound as a length.
3. Option C, \( \mu h \): this is exactly what the torque balance about the top point yields.
4. Option D, \( \mu^2 h^2 \): this is not even dimensionally a length in the way the problem needs, and does not follow from any term in the torque equation.

The torque balance identifies option C as correct.

Therefore, the correct answer is \( \mu h \).