Answer to Question #203833 in Civil and Environmental Engineering for Dion

Question

Consider a 7.2 m ladder with a painter climbing up it. If the mass of the ladder is 15.0 kg, the mass of the painter is 83.2 kg, and the ladder begins to slip at its base when his feet are 39% of the way up the length of the ladder, what is the coefficient of static friction between the ladder and the floor? μwall = 0.265. The angle between the floor and ladder is 41°. Please ensure that all unknown answers are given. g=9.81m/s2. Find N1 and N2 as well as µ

Answer 1

$\beta_1=\mu N_1$

$\beta_2=\mu _{wall}N_2=0.265N_2$

Equilibrium m vertical direction gives.

$\beta_2+N_1=(m+m)g$

Equilibrium in horizontal direction,

$N_2=\beta,$

$=N_2=\mu N,$

$=0.265 \mu N,+N_1=(m+m)g$

$=0.265 \mu N, +N_1=1088.91$

Since , the system is stable.

$\therefore \sum C_O=0$

$=mg*0.39L cos45^0+mg*0.5L cos 41^0-N_2L sin41^0- \beta_2L cos41^0=0$

$L(268.53+66.63-0.656 \mu N,-0.200 \mu N_1)=0$

$=335.16=0.856 \mu N,$

$=\mu N, = 391.54$

$N_1=985.15$

$\mu=\frac{\mu N_1}{N_1}=0.3974$