A uniform ladder whose length is$5.0m$and whose weight is$400N$leans against a frictionless vertical wall. The coefficient of static friction between the level ground and the foot of the ladder is$0.46$. What is the greatest distance the foot of the ladder can be placed from the base of the wall without the ladder immediately slipping?

By drawing the free body diagram of the ladder, we can find the forces acting in x and y directions.By taking the summation of moments at the foot of the ladder, we can determine the greatest distance between the foot of the ladder and the base of the wall.

Formula:

${\sum }_{}^{}{\text{F}}_{\text{y}}=0$

${\sum }_{}^{}{\text{F}}_{\text{x}}=0$

${\sum }_{}^{}{\text{M}}_{\text{footoftheladder}}=0$

${\text{F}}_{\text{staticfriction}}={\text{μ}}_{\text{s}}\times \text{NormalForce}$

From this,

${\sum }_{}^{}{\text{F}}_{\text{y}}=0$

${\text{F}}_{\text{n}}-\text{mg}=0$

${\text{F}}_{\text{n}}=\text{mg}$

${\sum }_{}^{}{\text{F}}_x=0$

$\begin{array}{c}{\text{F}}_{\text{W}}-{\text{F}}_{\text{S}}=0\\ {\text{F}}_{\text{W}}={\text{F}}_{\text{S}}\end{array}$

We have,

${\text{F}}_{\text{staticfriction}}={\text{μ}}_{\text{s}}\text{×NormalForce}$

${\mu }_s=\frac{F_{staticfriction}}{NormalForce}$

${\text{μ}}_{\text{s}}=\frac{{\text{F}}_{\text{w}}}{\text{mg}}$

We take summation of the moment due to all forces at the base of the ladder,

${\sum }_{}^{}{M}_{footoftheladder}=0$

$\begin{array}{c}-(\text{mg}\times \frac{\text{a}}{\text{2}})+({\text{F}}_{\text{w}}\times \text{h})=0\\ (\text{mg}\times \frac{\text{a}}{\text{2}})={\text{F}}_{\text{w}}\times \text{h}\\ \frac{\text{a}}{2h}=\frac{{\text{F}}_{\text{w}}}{mg}\\ Thus,\\ {\mu }_s=\frac{{\text{F}}_{\text{w}}}{mg}=\frac{\text{a}}{2h}\end{array}$

From the diagram, we can say that,

$\text{L}=\sqrt{{\text{h}}^{\text{2}}+{\text{a}}^{\text{2}}}$

${\text{L}}^{\text{2}}={\text{h}}^{\text{2}}+{\text{a}}^{\text{2}}$

$\text{h}=\sqrt{{\text{L}}^2-{\text{a}}^{\text{2}}}$

We put the value of h in the equation of moment,

${\mu }_s=\frac{\left(\frac{a}{2}\right)}{\sqrt{L^2-a^2}}$

$({\mu }_s\times \left(\sqrt{L^2-a^2}\right))=\left(\frac{a}{2}\right)$

Taking square at both sides,

${\text{μ}}_{\text{s}}^{\text{2}}\times ({\text{L}}^{\text{2}}-{\text{a}}^{\text{2}})=\frac{{\text{a}}^{\text{2}}}{\text{4}}$

$\begin{array}{c}({\text{μ}}_{\text{s}}^{\text{2}}\times {\text{L}}^{\text{2}})-({\text{μ}}_{\text{s}}^{\text{2}}\times {\text{a}}^{\text{2}})=\frac{{\text{a}}^{\text{2}}}{\text{4}}\\ ({\text{μ}}_{\text{s}}^{\text{2}}\times {\text{L}}^{\text{2}})=\frac{{\text{a}}^{\text{2}}}{\text{4}}+({\text{μ}}_{\text{s}}^{\text{2}}\times {\text{a}}^{\text{2}})\\ ({\text{μ}}_{\text{s}}^{\text{2}}\times {\text{L}}^{\text{2}})={\text{a}}^{\text{2}}\times (\frac{\text{1}}{\text{4}}+{\text{μ}}_{\text{s}}^{\text{2}})\\ ({\text{μ}}_{\text{s}}^{\text{2}}\times {\text{L}}^{\text{2}})={\text{a}}^{\text{2}}\times \frac{\left({\text{1+4μ}}_{\text{s}}^{\text{2}}\right)}{\text{4}}\\ \frac{\text{4}\times ({\text{μ}}_{\text{s}}^{\text{2}}\times {\text{L}}^{\text{2}})}{(\text{1}+{\text{4μ}}_{\text{s}}^{\text{2}})}={\text{a}}^{\text{2}}\end{array}$

$\begin{array}{c}\text{a}=\sqrt{\frac{\text{4×}\left({\text{μ}}_{\text{s}}^{\text{2}}{\text{×L}}^{\text{2}}\right)}{\left({\text{1+4μ}}_{\text{s}}^{\text{2}}\right)}}\\ =\sqrt{\frac{4\times ({0.46}^2\times {\left(5.0m\right)}^2)}{(1+4\times {0.46}^2)}}\\ =3.38m\\ \approx 3.4m\end{array}$

The greatest distance between the foot of the ladder and the base of the wall without the ladder immediately slipping should be$3.4m$