Question: A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weighs 160 N. The coefficient of static friction between the foot of the ladder and the ground is 0.40. A man weighing 740 N climbs slowly up the ladder. Start by drawing a free-body diagram of the ladder.

(a) What is the maximum friction force that the ground can exert on the ladder at its lower end?

(b) What is the actual friction force when the man has climbed 1.0 m along the ladder?

(c) How far along the ladder can the man climb before the ladder starts to slip?

The condition for translational equilibrium is: $\mathbf{\sum }{\mathit{F}}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{=}\mathbf{0}$and that for rotational equilibrium is: $\mathbf{\sum }{\mathit{\tau }}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{=}\mathbf{0}$. The total of the forces will be zero.

The ladder, placed against a frictionless wall, is at equilibrium. The wall exerts a normal force of $N_1\text{Newton}$, and the normal force exerted by the ground is $N_2\text{Newton}$..

Let the whole given setup be illustrated as a free body diagram for forces on the ladder, as shown in the figure as:.

Here, the weight of the ladder and that of man are, respectively $W_{I}and{W}_M$..

Now, the frictional force applied by the ground on the ladder will be:

$$\begin{gathered} F={\mu }_s{N}_2 \\ F=0.40N_2 \end{gathered}$$ ...........(i)

Since the ladder is at equilibrium, the weights and the forces will be related as:

$$\begin{gathered} N_2=W_l+W_m \\ =160+740 \\ =900\text{N} \end{gathered}$$

Put this value in equation (i), we get:

$$\begin{gathered} F=0.40N_2 \\ =0.40\times 900 \\ =360\text{N} \end{gathered}$$

Thus, the friction force applied by the ground is $360\text{N}$.

Applying the condition for rotational equilibrium, the torque on the ladder can be calculated as:

$$\begin{gathered} \sum \tau =0 \\ {F}_{\text{wall}}·\left(4\right)=\left(1.5\right)W_l+\left(1\right){\mu }_s{W}_m \\ 4F_{\text{wall}}=1.5\times 160+1\times 0.6\times 740 \\ {F}_{\text{wall}}=171\text{N} \end{gathered}$$

Thus, the friction force when the man had climbed 1 meter was 171 Newtons.

Assuming that the man has climbed the distancexmeters, the torque just before the ladder started to slip can be calculated as:

$$\begin{gathered} \sum \tau =0 \\ {N}_2\left(4\right)=\left(1.5\right)W_l+\left(x\right){\mu }_s{W}_m \\ 4\times 360=1.5\times 160+\left(x\right)\times 0.6\times 740 \\ x=2.7\text{m} \end{gathered}$$

Thus, the ladder started to slip when the man climbed the ladder to the length of 2.7 meters.