Question

(1) A 4m ladder has mass m= 10 kg. We assume that its centre of mass is half-way along its length. The ladder is leaning against a wall at an angle of 30° from the vertical, as shown in the figure. Suppose that the wall is frictionless smooth and slippery), and that the coefficient of static friction between the feet of the ladder and the floor is H. =0.6. Take the acceleration due to gravity to be g=10m/s" directed vertically downward. We will answer the question whether a person with mass M = 90 kg can safely stand on the ladder at its 3/4 point. (a) Draw the following forces on the diagram (on the right) showing the ladder. Label the forces and indicate the direction of each force with an arrow from the point where it acts. (i) the weight of the ladder, mg. (ii) the weight of the person, Mg. (ii) the normal force of contact between the floor and the feet of the ladder, 30 (iv) the force of static friction between the floor and the feet of the ladder, f (v) the normal force of contact between the wall and the head of the ladder, (b) Sum the forces in the vertical direction to determine the magnitude of the normal force acting at the floor. (c) Compute the maximum possible value of the force of static friction acting between the floor and the ladder, .max = . N. (d) Choose the torque point (axis) to be where the ladder's feet rest on the floor. Compute the torque produced by (i) the weight of the ladder. (ii) the weight of the person. (ii) the normal force of contact between the floor and the feet of the ladder. (iv) the force of static friction between the floor and the feet of the ladder. (v) the normal force of contact between the wall and the head of the ladder. (e) Use the fact that the net torque must vanish to compute the magnitude of the normal force from the wall. (f) Infer the magnitude of the force of static friction required to hold the ladder in place. IS IS NOT safe. (g) The person standing on the ladder CHOOSE ONE:)

Answer 1

A. Nw - B IC Mſ N. I mg - 145 D

@ An the forces and their directions are drawn in the above ilustration. Green dot is the mark where the person stands (marked point B) A Red dot is the center of gravity of the ladder from where the weight of the ladder acts. (marked point c) A is the contact point of the ladder and the wall. D is the contact point of the ladder and the floor. o is the wall a floor connection point The ladder is 30° tilted from the vertical LADO (30–30) = 609 Ć BD AD =4m 3 x 4 =3 m 1 xy =2m CD 6 All the forces in the vertical directions ane-

Мо Г. и 10 - height of the ladder mg [acting down hand) → Weight person ] → Normal reaction force acting at the floor N. [acting upward] : Static force balance gives - mg + Mg Ng ma 10kg M = gokg g=10m m/sv ci (10810) + (90x10) Nt NH 1000 N o The The friction coefficient at the bottom us = 0.6 ::. Left ward acting static friction force fs, MSX NE 0.6 X 1000 ts, 600 M

@ As indicated the torque point is - We know Torque = Force x smallest distance / perpendicular distance between the torque point and the line of action of the force. is Horizontal distance between the torque point (D) and the ladder weight CDH CD cos 60 & Torane due to ladder weight T mg x CDH 100 X 1 = +100 N-m O Anticlockwise torque tre Clock wise torque -) - ve w Horizontal distance between the person and the torque point (D) BDH BD Cos 60 1.5m :: Torane due to the person weight. T. Р Mg x BDH =+1350 M-m

(im The distance between floor normal force (NA) and the torq ne point D is on Torque due to due to Ng =0 / Ti = N, хо (iv) Simimilerly as the previous one the distance between the friction force at floor and the torque point D is ou : Torque due to fs = 0 F, хо ff & Vertical distance between the normal reaction force of wall at Ą and the torque point D- AD sin 60 = 3.4641 : Torque due to Nw Nam TH - Nw X 3.4641 [ure sign, because the torque is clock wise @ For static condition, net torque about the torque point must be equal to on

Tht Tp + Tj + THt Tw = 0 100 + 1350 +0 40 - 3.4641 MW Nw 418.6 N Am For static condition all the horizontal forces must balance out. . ut Nw =418.6 N . Required magnetude of static friction at the floor = 418.6 N ts = AN 6 Since required <ts, The person on the ladder is safe If you have any confusion with the solution, ask me in the comments.