A$\mathbf{75}\mathbf{}\mathit{k}\mathit{g}$window cleaner uses a$\mathbf{10}\mathbf{}\mathit{k}\mathit{g}$ladder that is$\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}$long. He places one end on the ground$\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathit{m}$from a wall, rests the upper end against a cracked window, and climbs the ladder. He is$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}$up along the ladder when the window breaks. Neglect friction between the ladder and window and assume that the base of the ladder does not slip. When the window is on the verge of breaking, what are (a) the magnitude of the force on the window from the ladder, (b) the magnitude of the force on the ladder from the ground, and (c) the angle (relative to the horizontal) of that force on the ladder?

Mass of thewindow cleaner$M=75kg$

Mass of the ladder$m=10.0kg$

Length of the ladder$L=5.0m$

Distance from the foot of the ladder to theposition of the window cleaner$\ell =3.0m$

Distance from the wall to the foot of the ladder$d=2.5m$

The forces on the ladder are shown in the diagram$F_1$is the force of the window, horizontal because the window is frictionless.$F_2$and$F_3$are the components of the force of the ground on the ladder. MIs the mass of the window cleaner andis the mass of the ladder.

The man is affected by gravity at a point $3.0m$up the ladderand the centre of the ladder experiences the effects of gravity.Let $\theta$be the angle between the ladder and the ground.

$$\begin{gathered} \backslash \mathrm{begingathered}cos\theta =\frac{d}{L} \\ \theta =co{s}^{-1}\left(\frac{2.5}{5}\right) \\ ={60}^0 \end{gathered}$$

(a)

Since the ladder is in equilibrium the sum of the torques about its foot or any other point vanishes.

Let $\ell$be thedistance from the foot of the ladder to the position of the window cleaner.

$$\begin{gathered} Mg\ell cos\theta +mg\frac{L}{2}cos\theta -F1Lsin\theta =0 \\ F1=\frac{(M\ell +m\frac{L}{2})gcos\theta }{Lsin\theta } \\ =\frac{\left[(75 kg)(3.0 m)+(10 kg)(2.5 m)(9.8 m/s^2)cos 60°\right]}{(5.0 m) sin 60°} \\ =2.8\times {10}^{2}N \end{gathered}$$

The magnitude of the force on the window from the ladderis.$2.8\times {10}^{2}N$

(b)

The sum of the horizontal forces and the sum of the vertical forces also vanish.

$$\begin{gathered} F1-F3=0 \\ F2-Mg-mg=0 \\ F1=F3=2.8\times {10}^{2}N \\ F2=(M+m)g \\ =(75kg+10kg)(9.8m/s^2) \\ =8.3\times {10}^{2}N \end{gathered}$$

The magnitude of the force of the ground on the ladder is given by the square root of the sum of the squares of its components

$\begin{array}{rcl}F& =& \sqrt{F{2}^2+F{3}^2}\\ & =& \sqrt{{(2.8\times {10}^{2}N)}^2+{(8.3\times {10}^{2}N)}^2}\\ & =& 8.8\times {10}^{2}N\\ & & \end{array}$

The magnitude of the force on the ladder from the ground is$8.8\times {10}^{2}N$

(c)

The angle $\varphi$between the force and the horizontal is given by

$\begin{array}{rcl}tan\varphi & =& \frac{F3}{F2}\\ & =& \frac{\left(830N\right)}{\left(280N\right)}\\ & =& 2.94\\ \varphi & =& 71°\\ & & \\ & & \end{array}$

The angle (relative to the horizontal) of that force on the ladder is$71°$