Question

A ladder is leaning against a vertical wall, and both ends of the ladder are at the point of slipping. The coefficient of friction between the ladder and the horizontal surface is ?1 = 0.215 and the coefficient of friction between the ladder and the wall is ?2 = 0.203. Determine the maximum angle with the vertical the ladder can make without falling on the ground.

Answer 1

Suppose that at the point of slipping the normal reactions at the ground and the wall are N1, N2 respectively, and the corresponding frictional forces are
R1 ? ?1N1 ... (1) and
R2 ? ?2N2 ... (2).
The inequalities occur because the coefficients are the limiting values for static friction - ie the friction force can be anything up to the limiting value ?N. At the point of slipping one of these will be an equality, but we do not know which at this stage.

Suppose the weight of the ladder is W. The ladder is in static equilibrium under these forces so
N1+R2 = W ... (3)
R1 = N2 ....(4)

Suppose the ladder has length 2L and makes angle A with the vertical. Take moments about the point of contact of the ladder with the wall. The ladder is not turning so the net moment of forces on it must be zero :
WLsinA + 2LR1cosA - 2LN1sinA = 0
W = 2(N1-R1cotA) ...(5)

Now it is a matter of eliminating W, N1, N2, R1, R2.

Substitute for W from (5) into (3) :
N1+R2 = 2(N1-R1cotA)
R2 = N1 - 2R1cotA.

Substitute from (4) and (2) :
R2 = N1 - 2N2cotA ? ?2N2
so
N1 ? ?2N2 + 2N2cotA
N1/N2 ? ?2 + 2cotA.

From (1) and (4) :
R1 = N2 ? ?1N1
1/?1 ? (N1/N2).

Therefore
1/?1 ? ?2 + 2cotA
2cotA ? (1/?1 )- ?2 = 4.44816
cotA ? 2.22408

TanA>= 0.4496
A ? 24.21 degrees.
(The inequality sign changes direction in the last line because as A increases cotA decreases.)

Hope it will help you.

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