A uniform ladder is 10 m long and weighs $\mathbf{200}\mathbf{\text{\hspace{0.17em}N}}$ . In Fig. 12-78, the ladder leans against a vertical, frictionless wall at height$\mathbf{h}\mathbf{}\mathbf{=}\mathbf{8}\mathbf{.0}\mathbf{\text{\hspace{0.17em}m}}$ above the ground. A horizontal force is applied to the ladder at distance $\mathbf{d}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.0}\mathbf{\text{\hspace{0.17em}m}}$from its base (measured along the ladder).

(a) If force magnitude$\mathbf{F}\mathbf{}\mathbf{=}\mathbf{50}\mathbf{\text{\hspace{0.17em}N}}$ , what is the force of the ground on the ladder, in unit-vector notation?

(b) If$\mathbf{F}\mathbf{}\mathbf{=}\mathbf{150}\mathbf{\text{\hspace{0.17em}N}}$ , what is the force of the ground on the ladder, also in unit-vector notation?

(c) Suppose the coefficient of static friction between the ladder and the ground is$\mathbf{\text{0.38}}$ for what minimum value of the force magnitude Fwill the base of the ladder just barely start to move toward the wall?

The length of the ladder,$\text{L}=10\mathrm{m}$

The weight of the ladder,$\text{W}=200\text{\hspace{0.17em}N}$

The height of the wall,$\text{h}=8.0\text{\hspace{0.17em}m}$

The distance from the base of the ladder at which$\overrightarrow{\mathrm{F}}$is applied as$\text{d}=2.0\text{\hspace{0.17em}m}$

The magnitude of the applied force,$\text{F}=50\text{\hspace{0.17em}N}$

The coefficient of static friction, $\mathrm{\mu }=0.38$

You draw the free body diagram. The system is at equilibrium, for such a system, the vector sum of the forces acting on it is zero. The vector sum of the external torques acting on the ladder at one point is zero. The equations are given below.

$\begin{array}{c}\mathbf{\Sigma }{\overrightarrow{\mathbf{F}}}_{\mathbf{net}}\mathbf{=}\mathbf{0}\\ \mathbf{\Sigma }{\overrightarrow{\mathbf{\tau }}}_{\mathbf{net}}\mathbf{=}\mathbf{0}\end{array}$

You can draw the free body diagram for the ladder as shown in the figure. You choose the axis of rotation passing through the point $\mathrm{O}$ at the top of the ladder.

You can use the Pythagorean law and find the horizontal distance $\mathrm{x}$ as

$\begin{array}{c}\mathrm{x}=\sqrt{{\mathrm{L}}^2-{\mathrm{h}}^2}\\ =\sqrt{{\left(10\mathrm{m}\right)}^2-{\left(8.0\mathrm{m}\right)}^2}\\ =6.0\text{\hspace{0.17em}m}\end{array}$

The line of action of$\mathrm{F}$has the moment of the arm as$\mathrm{r}$then

$\begin{array}{c}\mathrm{r}=(\mathrm{L}-\mathrm{d})\sin \mathrm{\theta }\\ =(\mathrm{L}-\mathrm{d})\frac{\mathrm{h}}{\mathrm{L}}\\ =(10\text{m}-2.0\text{m})\frac{8.0\text{m}}{10\text{m}}\\ =6.4\text{\hspace{0.17em}m}\end{array}$

According to the static equilibrium condition, the sum of the vertical forces acting on the ladder is zero. Hence,

$\begin{array}{c}{\mathrm{\Sigma F}}_{\mathrm{y},\mathrm{net}}=0\\ {\mathrm{F}}_{\mathrm{g},\mathrm{y}}-\mathrm{W}=0\\ {\mathrm{F}}_{\mathrm{g},\mathrm{y}}=\mathrm{W}=200\mathrm{N}\end{array}$

For the static equilibrium condition, the vector sum of the external torque acting on the ladder at about any point is zero.

$\begin{array}{c}{\mathrm{\Sigma \tau }}_{\mathrm{net}}=0\\ \mathrm{F}\times \mathrm{r}+\mathrm{W}\left(\frac{\mathrm{x}}{2}\right)+{\mathrm{F}}_{\mathrm{g},\mathrm{x}}\times \mathrm{h}-{\mathrm{F}}_{\mathrm{g},\mathrm{y}}\times \mathrm{x}=0\left(1\right)\end{array}$

For $\mathrm{F}=50\text{\hspace{0.17em}N}:$

$\begin{array}{c}\mathrm{F}\times \mathrm{r}+\mathrm{W}\left(\frac{\mathrm{x}}{2}\right)+{\mathrm{F}}_{\mathrm{g}\mathrm{x}}\times \mathrm{h}-{\mathrm{F}}_{\mathrm{g}\mathrm{y}}\times \mathrm{x}=0\\ {\mathrm{F}}_{\mathrm{g}\mathrm{x}}\times \mathrm{h}=-\mathrm{F}\times \mathrm{r}-\mathrm{W}\left(\frac{\mathrm{x}}{2}\right)+{\mathrm{F}}_{\mathrm{g}\mathrm{y}}\times \mathrm{x}\\ {\mathrm{F}}_{\mathrm{g}\mathrm{x}}=\frac{-\mathrm{F}\times \mathrm{r}-\mathrm{W}\left(\frac{\mathrm{x}}{2}\right)+{\mathrm{F}}_{\mathrm{g}\mathrm{y}}\times \mathrm{x}}{\mathrm{h}}\left(2\right)\\ {\mathrm{F}}_{\mathrm{g}\mathrm{x}}=\frac{-50\mathrm{N}\times 6.4\mathrm{m}-200\mathrm{N}\left(\frac{6.0\mathrm{m}}{2}\right)+200\mathrm{N}\times 6.0\mathrm{m}}{8.0\mathrm{m}}\\ {\mathrm{F}}_{\mathrm{g}\mathrm{x}}=35\text{\hspace{0.17em}N}\end{array}$

The unit vector notation of the force of the ground on the ladder for $\text{F}=50\text{\hspace{0.17em}N}$ is $\overrightarrow{{\mathrm{F}}_{\mathrm{g}}}=\left(35\text{\hspace{0.17em}N}\right)\widehat{\mathrm{i}}+\left(200\text{\hspace{0.17em}N}\right)\widehat{\mathrm{j}}$.

For$\text{F}=150\text{\hspace{0.17em}N}$

From equation (2)

$\begin{array}{c}{\mathrm{F}}_{\mathrm{g}\mathrm{x}}=\frac{-150\mathrm{N}\times 6.4\mathrm{m}-200\mathrm{N}\left(\frac{6.0\mathrm{m}}{2}\right)+200\mathrm{N}\times 6.0\mathrm{m}}{8.0\mathrm{m}}\\ =-45\text{\hspace{0.17em}N}\end{array}$

The unit vector notation of the force of the ground on the ladder for $\text{F}=150\text{\hspace{0.17em}N}$ is$\overrightarrow{{\mathrm{F}}_{\mathrm{g}}}=(-45\text{\hspace{0.17em}N})\widehat{\mathrm{i}}+\left(200\text{\hspace{0.17em}N}\right)\widehat{\mathrm{j}}$.

The minimum value of the force with the base of the ladder just barely starts to move towards the wall.

It implies that the frictional force is acting along the negative x-axis in between the surfaces of the ladder and the earth.

For this now,${\text{F}}_{\text{N}}$is the normal force acting on the base of the ladder, then equating the vertical forces acting on it as

${\mathrm{F}}_{\mathrm{N}}={\mathrm{F}}_{\mathrm{g}\mathrm{y}}=200\text{\hspace{0.17em}N}$

For horizontal forces,

$\mathrm{f}=-{\mathrm{F}}_{\mathrm{g}\mathrm{x}}$

According to the definition of the static frictional force,

$\begin{array}{c}\mathrm{f}={\mathrm{\mu }}_{\mathrm{s}}{\mathrm{F}}_{\mathrm{N}}\\ =0.38\times 200\text{\hspace{0.17em}N}\\ =76\text{\hspace{0.17em}N}\end{array}$

From equation (1) as

$\begin{array}{c}\mathrm{F}=\frac{-\mathrm{W}\left(\frac{\mathrm{x}}{2}\right)-{\mathrm{F}}_{\mathrm{g}\mathrm{x}}\times \mathrm{h}+{\mathrm{F}}_{\mathrm{g}\mathrm{y}}\times \mathrm{x}}{\mathrm{r}}\\ =\frac{-200\mathrm{N}\left(\frac{6.0\mathrm{m}}{2}\right)-(-76\mathrm{N})\times (8.0\mathrm{m})+(200\mathrm{N})\times (6.0\mathrm{m})}{6.4\mathrm{m}}\\ =1.9\times {10}^2\text{\hspace{0.17em}N}\end{array}$