Question

If the coefficient of static friction is 0.60, and the same ladder makes a 53° angle with respect to the horizontal, how far along the length of the ladder can a 63.0 kg person climb before the ladder begins to slip? The ladder is 8.2 m in length and has a mass of 18 kg.

mg A uniform ladder at rest leaning against a smooth wall.

Answer 1

P = 0.60x(18+63)x9.8 =476.28 N

$Plsin\theta -mgl/2cos\theta -m'gxcos\theta =0 where m= mass of ladder m'= mass of person l= length of ladder x= distance upto the person can climb before slipping$ $Plsin\theta -mgl/2cos\theta -m'gxcos\theta =0 where m= mass of ladder m'= mass of person l= length of ladder x= distance upto the person can climb before slipping x = (Plsin\theta -mgl/2cos\theta )/m'gcos\theta$

X= (476.28x8.2xsin53⁰ - 18x9.8x4.1xcos 53⁰) / 63x9.8xcos53⁰

^{= 7.22 m}