A uniform 10 kg ladder leans at 60° to the floor against a smooth wall. Find the floor friction and the wall reaction — Physics problems | PhysLabo
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A ladder against a smooth wall: friction and reaction forces

Medium 10 grade Mechanics / Statics
Problem statement

A uniform ladder of mass $m = 10$ kg and length $L$ leans against a smooth vertical wall and a rough floor, making an angle $\alpha = 60°$ with the floor (see the figure). Find the reaction force $N_w$ of the wall and the friction force acting on the ladder from the floor, as well as the minimum coefficient of friction of the floor for which the ladder does not slip. Take $g = 10$ m/s², $\tan 60° \approx 1.73$.

A uniform ladder of length L leans on a smooth vertical wall and a rough floor, making angle α with the floor
Given
  • $m = 10$ kg
  • $\alpha = 60°$, $\tan 60° \approx 1.73$
  • $g = 10$ m/s²
  • $N_w = ?$, $F_{fr} = ?$, $\mu_{min} = ?$
Hints
Hint 1 / 2
Take the axis of torques at the bottom of the ladder: two unknown forces ($N$ and $F_{fr}$) pass through it, so their torques are zero.
Hint 2 / 2
The lever arm of gravity is $\frac{L}{2}\cos\alpha$, that of the wall reaction is $L\sin\alpha$. The wall is smooth, so its reaction is perpendicular to the wall.
Solution

The forces on the ladder are gravity $m\vec g$ at its midpoint, the wall reaction $\vec N_w$ (horizontal, since the wall is smooth), the floor reaction $\vec N$ (upward) and friction $\vec F_{fr}$ (along the floor, toward the wall) at the bottom point $O$. The ladder is in equilibrium: the net force and the net torque are zero.

Forces on the ladder and their lever arms about point O
Forces on the ladder and their lever arms about point O

1. Vertical and horizontal forces:

$$N = mg = 100 \text{ N}, \qquad F_{fr} = N_w$$

2. Torques about point $O$ (so $N$ and $F_{fr}$ drop out). The lever arm of gravity is $\frac{L}{2}\cos\alpha$, that of $N_w$ is $L\sin\alpha$:

$$N_w\,L\sin\alpha = mg\,\frac{L}{2}\cos\alpha \;\Rightarrow\; N_w = \frac{mg}{2\tan\alpha} = \frac{100}{2 \cdot 1.73} \approx 28.9 \text{ N}$$

3. Friction is $F_{fr} = N_w \approx 28.9$ N. For the ladder not to slip we need $F_{fr} \le \mu N$:

$$\mu_{min} = \frac{F_{fr}}{N} = \frac{28.9}{100} \approx 0.29$$
Answer

$N_w = F_{fr} \approx 28.9$ N; $\mu_{min} \approx 0.29$

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