Question

(1). A uniform ladder ofmass m = 40 Kg, and length I:Dmis leaned against a smooth vertical wall. A person of mass Masss M-0 Kg smands on tue latidera distance罢= 7m fom the bottom, as measured along the ladder The foot ofthe ladder is d1.2mfrom the botom of tue wall Whhat is the fonpe enemed by te wall on the ladder? What is the normmal forpe enenet bw the foor on the laider persan 2 ME geund

Answer 1

Given, from the figure s - normal reaction at the wall R - normal reaction at the ground f - frictional force exerted by the ground on d - foot of the ladder m - mass of the ladder m - mass of the person the angle by the ladder the ground satisfies theta = cos^- 1(d/l) = cos^- 1(1 2/10) theta = 83.11 degree the reaction force acts to twist the ladder is the opposite sense hence setting net torque to zero we get mg 1/2 cos theta + mg cos theta - sin theta = 0 substituting the given information and solving for s, we get s = (mg/2 + mg x/l)/tan theta = (0. times 40 times 9.81) + (80 times 9.81 times 7/10)/tan(83.11 degree) s = 90.09 n Therefore the force exerted by the wall on the ladder is 90.09 N.