A very thin ladder of length L and mass M leans against a vertical wall, on a ho

Question

A very thin ladder of length L and mass M leans against a vertical wall, on a horizontal floor, making an angle of with respect to the wall. Imagine that there is a large coefficient of friction at the floor so that the ladder is in static equilibrium. but assume that the wall is frictionless.

a) Draw a free body diagram for the ladder, showing all forces acting.

b) Using the bottom of the ladder as the axis of rotation or origin compute all the forces and torques on the ladder such that it is in equilibrium.

c) why did I make the wall frictionless?

d) re-solve the problem using the TOP of the ladder as the axis of rotation or origin. What is different in the end?

e) At what angles would the ladder start to slip? if = 0.8 (not unreasonable for rubber ladder feet on the wood floor), what is the maxium angle at which you could lean the ladder?

Please help me solve this questions. THANK YOU!

Explanation / Answer

(a) (b) writing equations along X and Y direction N1 = Mg f=µN1 N2=f solving the equations we get N1=Mg , f = N2 = µMg torque due to Mg = MgLsin?/2 torque due to N2 = µMgLcos? torque due to other forces are zero (c) if the wall is with friction then it would be difficult to compute other values of the forces (d) torque due to Mg = MgLsin?/2 torque due to N1 = MgLsin? torque due to friction = MgLcos? (e) he lader should be in rotational equilibrium also so considering bottome nd as origin torque due to Mg = torque due to N2 MgLsin?/2 = µMgcos? => tan? = 2µ given µ = 0.8 => ? = arc tan (1.6) = 57.994 degrees please rate livesaver

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