91Ó°ÊÓ

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:

${∑}_{}^{}{\text{F}}_{\text{y}}=0$

${∑}_{}^{}{\text{F}}_{\text{x}}=0$

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

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

From this,

${∑}_{}^{}{\text{F}}_{\text{y}}=0$

${\text{F}}_{\text{n}}−\text{mg}=0$

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

${∑}_{}^{}{\text{F}}_x=0$

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

We have,

${\text{F}}_{\text{staticfriction}}={\text{μ}}_{\text{s}}\text{×±·´Ç°ù³¾²¹±ô¹ó´Ç°ù³¦±ð}$

${\mathrm{μ}}_s=\frac{F_{staticfriction}}{NormalForce}$

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

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

${∑}_{}^{}{M}_{footoftheladder}=0$

$\begin{array}{c}−(\text{mg}×\frac{\text{a}}{\text{2}})+({\text{F}}_{\text{w}}×\text{h})=0\\ (\text{mg}×\frac{\text{a}}{\text{2}})={\text{F}}_{\text{w}}×\text{h}\\ \frac{\text{a}}{2h}=\frac{{\text{F}}_{\text{w}}}{mg}\\ Thus,\\ {\mathrm{μ}}_s=\frac{{\text{F}}_{\text{w}}}{mg}=\frac{\text{a}}{2h}\end{array}$

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,

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

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

Taking square at both sides,

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

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

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

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