Question: In Fig. 12-7and the associated sample problem, let the coefficient of static friction${\mu }_s$ between the ladder and the pavement is 0.56 . How far (in percent) up the ladder must the firefighter go to put the ladder on the verge of sliding?

The coefficient of static friction ${\mu }_s$between the ladder and the pavement is 0.53

As per Newton’s second law, the force acting on the object is directly proportional to the acceleration of the object. The net force on the object is the vector sum of all the forces acting on the object.

Using the formula for Newton’s second law to the vertical and horizontal direction and condition for static equilibrium, write the equation for force and torque about point O. Solving these equations, we can find how far (in percent) up the ladder must the firefighter go to put the ladder on the verge of sliding.

Static Equilibrium conditions:

$$\begin{gathered} \sum F_x=0 \\ \sum F_y=0 \\ \sum \tau =0 \end{gathered}$$

Let be the horizontal distance between the origin O and the firefighter just before the ladder is on the verge of sliding.

$F_w$is the horizontal force from the wall and $F_{py}$is the vertical component. $F_{px}$is the horizontal component of the force, $\overrightarrow{F_p}$on the ladder from the pavement

Let Mg and mg be the downward gravitational forces on the firefighter and the ladder respectively.

Consider according to Newton’s second law to the vertical and horizontal directions, the net force F_net in the x, and y direction can be written as,

$$\begin{gathered} \sum F_{net},x=F_w-F_{px}=0 \\ \sum F_{net},y=F_{py}-\left(M+m\right)g=0 \end{gathered}$$

When the ladder is on the verge of sliding,

,$$\begin{gathered} F_{px}=\mu F_{py} \\ {F}_w=F_{px}=\mu \left(M+m\right)g \end{gathered}$$

The net torque about point O must be, just before sliding

$$\begin{gathered} \sum _0{\tau }_{net}=-h\left(F_w\right)+x\left(Mg\right)+\left(mg\right) \\ =0 \end{gathered}$$

Ladder’s center of mass is at L/3 as described in the sample problem 12.03. As shown in figure, weight of the ladder would be at distance a/3 from the point O.

$$\begin{gathered} x=\frac{h{F}_w-\left(\frac{a}{3}\right)mg}{Mg} \\ =\frac{h{\mu }_s\left(M+m\right)g-\left(\frac{a}{3}\right)mg}{Mg} \\ =\frac{h{\mu }_s\left(M+m\right)-\left(\frac{a}{3}\right)m}{M} \end{gathered}$$

Divide both sides of the equation obtained in step 5 by a and solve:

$$\begin{gathered} \frac{x}{a}=\frac{h{\mu }_s\left(M+m\right)-\left(\frac{a}{3}\right)m}{Ma} \\ =\frac{\left(9.3 m\right)\left(0.53\right)\left(72kg+45 kg\right)-\left(\frac{7.58m}{3}\right)\left(45kg\right)}{\left(72kg\right)\left(7.58m\right)} \\ =0.848=85\% \end{gathered}$$

The firefighter must go 85% up the ladder to put the ladder on the verge of sliding.