(III) A uniform ladder of mass mand length leans at an angle\(\theta \)against a frictionless wall, Fig. 9–70. If the coefficient of static friction between the ladder and the ground is\({\mu _s}\). Determine a formula for the minimum angle at which the ladder will not slip.

The given data can be listed below as:

• The length of the ladder is l.
• The angle of inclination of the ladder with the ground is\(\theta \).
• The acceleration due to gravity is\(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).
• The coefficient of static friction is\({\mu _s}\).

From Newton’s second law, the forces along the x and y directions can be calculated. The net acceleration of the ladder becomes zero.Thus, the net force on the ladder in the x and y directions should be equal to zero. At equilibrium, the net torque acting on the ladder should be zero.

With the help of these equilibrium equations, the minimum inclined angle can be evaluated.

The diagram of the ladder can be shown as:

Here, Mg is the weight of the ladder,\(\theta \)is the angle of inclination of the ladder with the ground, which is also considered as the angle at which the ladder will not slip,\({F_{NG}}\)is the normal force on the ladder due to the ground,\({F_{{\rm{fr}}}}\)is the frictional force,\({F_{{\rm{NW}}}}\)is the normal reaction due to the wall, lis the length of the ladder, and C is the point where the center of gravity of the ladder acts.

At equilibrium, the net forces acting in the vertical direction should be equal to zero.

\(\begin{array}{c}{\sum F _y} = 0\\{F_{NG}} - Mg = 0\\{F_{NG}} = Mg\end{array}\)

At equilibrium, the net forces acting in the horizontal direction should also be equal to zero.

\(\begin{array}{c}{\sum F _x} = 0\\ - {F_{NW}} + {F_f} = 0\\{F_{NW}} = {\mu _s}{F_{NG}}\\{F_{NW}} = {\mu _s}Mg\end{array}\)

At equilibrium, the net torques acting about point A should be zero. Therefore, the torques equation can be expressed as:

\(\begin{array}{c}\sum {{T_{net}}} = 0\\ - AB \times {F_{{\rm{NW}}}}\sin \theta + AC \times Mg\cos \theta = 0\\ - l \times {\mu _s}Mg\sin \theta + \frac{l}{2} \times Mg\cos \theta = 0\\l \times {\mu _s}Mg\sin \theta = \frac{l}{2} \times Mg\cos \theta \end{array}\)

This can be further solved as:

\(\begin{array}{c}2{\mu _s}Mg\sin \theta = Mg\cos \theta \\2{\mu _s}\sin \theta = \cos \theta \\\tan \theta = \frac{1}{{2{\mu _s}}}\\\theta = {\tan ^{ - 1}}\left( {\frac{1}{{2{\mu _s}}}} \right)\end{array}\)

Thus, the formula for the minimum angle at which the ladder will not slip is \(\left\{ {{{\tan }^{ - 1}}\left( {\frac{1}{{2{\mu _s}}}} \right)} \right\}\).