posted by Anon .
A ladder of length 2L and mass M is positioned on level ground leaning against a wall such that the angle between the ladder and the horizontal is α. The coefficient of static friction between the ladder and the wall and between the ladder and the ground is μstatic = 0.65. The centre of mass of the ladder is halfway along it.
(c) For the ladder to be in mechanical equilibrium:
(i) Write down equations for the total x- and y-components of the 5 forces acting on the
(ii) Consider torques about the centre of the ladder. In which direction (into or out of the page) does the torque due to each of the 5 forces force act?
(iii) Write down an equation for the sum of the torques about the centre of mass of the ladder.
(iv) Use your equation for the torques to derive an expression for tan α in terms of the magnitudes of the forces acting.
(d)(i) If the ladder is just on the point of slipping at both the upper and lower ends, what
can you say about the pair of forces acting at each of the top and bottom of the ladder?
(ii) Hence use this information, with the information from (c)(i) and the expression you have derived in (c)(iv) to calculate the minimum angle that the ladder can form with the ground in order for it not to slip.