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

Step 1: Since the ladder is light and the wall is smooth, the floor alone supports the man's weight: \( N_f = mg \), and at the verge of slipping the friction is \( f = \mu mg \).

Step 2: Horizontal equilibrium requires the wall's push to balance the friction: \( N_w = f = \mu mg \).

Step 3: The three forces on the ladder — the man's weight, the floor's reaction (normal plus friction), and the wall's reaction — must act through a common point for the ladder to stay in equilibrium. Since the floor's combined reaction makes an angle \( \arctan\mu \) with the vertical, and the wall's reaction acts horizontally at height \( h \), these two lines of action meet at a horizontal distance \( h\tan(\arctan\mu) = \mu h \) from the base.

Step 4: For equilibrium, the man's vertical weight must also pass through that same point, so his horizontal distance from the base is exactly \( d = \mu h \).

\[ \boxed{d = \mu h} \]

Answer 2

This approach uses the fact that a rigid body in equilibrium under exactly three forces must have all three forces meet at a single point (or all be parallel) — a standard result for three-force members.

The three forces on the ladder are: the man's weight \( mg \), acting straight down at his position; the wall's reaction \( N_w \), acting horizontally at the top (height \( h \)); and the floor's reaction, which combines the normal force \( N_f = mg \) and the friction \( f = \mu mg \) into a single resultant. This resultant makes an angle \( \phi \) with the vertical where \( \tan\phi = \frac{f}{N_f} = \mu \).

Extend the line of the wall's horizontal reaction (the horizontal line at height \( h \)) and the line of the floor's resultant reaction (starting at the base and tilted at angle \( \phi = \arctan\mu \) from the vertical). These two lines meet at a point whose horizontal distance from the base is \( h\tan\phi = h\mu \), since the resultant reaction line rises a height \( h \) while shifting horizontally by \( h\tan\phi \).

Because all three forces must be concurrent, the vertical line of the man's weight must also pass through this same intersection point. That means the man's horizontal distance from the base — the quantity the question asks for — equals \( \mu h \).

Testing the four options against this concurrency condition:

1. Option A, \( \mu^2 h \): would require the resultant floor reaction to tilt by an angle whose tangent is \( \mu^2 \), which contradicts \( \tan\phi = \mu \) coming directly from \( f = \mu N_f \).
2. Option B, \( \frac{\mu}{h} \): shrinks as \( h \) grows, which contradicts the geometry — a taller wall-contact point should shift the concurrency point further, not closer.
3. Option C, \( \mu h \): matches the horizontal shift \( h\tan\phi = \mu h \) exactly.
4. Option D, \( \mu^2 h^2 \): has the wrong dimensions and the wrong power of \( \mu \) for a distance derived from a simple angle-tangent relation.

Only option C is consistent with the concurrency condition.

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