Answer to Question #171711 in Mechanics | Relativity for Lea Marie Depalubos

Question

2. A 30 foot ladder weighing 100 lbs having its center of mass one-third of the way up from the bottom rests against a smooth wall so that it makes an angle of 60 degrees with the ground. If the coefficient of friction between the ground and the ladder is 0.4, how high can a 150-lb man go before the ladder slips. 

Answer 1

$\vec{F}+\vec{N_0}+m\vec{g}+M\vec{g}+\vec{N_l}=\vec{0},$

$\begin{cases} -F+N_l=0 \\ -mg-Mg+N_0=0 \end{cases}$

$\begin{cases} F=N_l\\ (m+M)g=N_0 \end{cases}$

$F\leqslant \mu N_0$

$\begin{cases} N_l\leqslant \mu N_0\\ (m+M)g=N_0 \end{cases}$

$N_l\leqslant \mu(m+M)g$

$mg\frac l3 cos \alpha +Mgxcos\alpha=N_l lsin\alpha$

$gcos\alpha (Mx+m\frac l3)=N_l lsin \alpha$

$Mx+m\frac l3=\frac{N_l ltan\alpha}{g} \leqslant\mu(m+M)l tan\alpha$

$x\leqslant l(\mu tan \alpha(1+\frac mM)-\frac{m}{3M})=27.97~ft.$