Answer to Question #157191 in Classical Mechanics for Yolande

Question

A uniform ladder AB of wieght W and length 2l rests with the end A against a smooth vertical wall and the end B on a rough horizontal ground. A man of weight equal to that of the ladder stands at the point C on the ladder, where BC = 5l/3 . The coefficient of friction between the ladder and the ground is 1/3. Given that the ladder is in limiting equilibrium when it makes an angle s to the horizontal;
(a) Show that tans = 2
(b) Find the magnitude of the reaction force at the wall and on the ground.

Answer 1

$\text {let }O\text{ middle of ladder}$

$AO=OB= l;AC=\frac{l}{3};BC=\frac{5l}{3}$

$s-\text{angle s to the horizontal;}$

$F_{fr}-\text{friction force at point }B$

$T_A-\text{wall reaction force}$

$T_B-\text{reaction force at point B}$

$mg - \text{the gravity of the ladder at a point }O$

$mg - \text{the gravity of the man at a point }C$

$\text{forces along the X-axis (horizontal)}$

$T_A-F_{fr}=0$

$\text{forces along the Y-axis (vertical)}$

$T_B-mg-mg=0;T_B-2mg=0;$

$\text{equilibrium condition about point }B$

$T_A\sin{(s)}-mg*\cos(s)*AC-mg*\cos(s)*AO=0$

$T_A\sin{(s)}-mg*\cos(s)*\frac{1}{3}-mg*\cos(s)*l=0$

$T_A=\frac{4}{3}\cot(s)mg$

$F_{fr}=\frac{4}{3}\cot(s)mg$

$a)\text{friction static of the ladder and the man} - \frac{1}{3}mg+\frac{1}{3}mg=\frac{2}{3}mg$

$\text{ limiting equilibrium} F_{fr} \text{equal to static friction}$

$F_{fr}=\frac{4}{3}\cot(s)mg$

$\frac{4}{3}\cot(s)mg=\frac{2}{3}mg$

$\cot(s)=\frac{1}{2}$

$\tan(s)=2$

$b)$ $T_A=F_{fr}=\frac{4}{3}\cot(s)mg$

$T_g-\text{reaction force on the ground.}$

$T_g= \sqrt{T_B^2+F_{fr}^2}$

$T_B=2mg$

$T_g= \sqrt{(4m^2g^2+\frac{16}{9}}\cot^2(s)m^2g^2)$

Answer:a)$\tan(s)=2$ b)$T_A=\frac{4}{3}\cot(s)mg$ ;$T_g= \sqrt{(4m^2g^2+\frac{16}{9}}\cot^2(s)m^2g^2)$