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 \( \mu h \).