(^152) A Textbook of Engineering Mechanics

*sin 45° = cos 45° = 0.707
Solution. Given: Length of the ladder (l) = 4 μ; Angle which the ladder makes with the
horizontal (α) = 45°; Coefficient of friction between the ladder and the wall (μw) = 0.4 and coefficient
of friction between the ladder and the floor (μf) = 0.5.
The forces acting on the ladder are shown in Fig. 9.4.
Let x = Distance between A and the man,
when the ladder is at the point
of slipping.
W = Weight of the ladder, and
Rf = Normal reaction at floor.
∴ Weight of the man
0.5
2
W
==W
We know that frictional force at the floor,
Ff = μf Rf = 0.5 Rf ...(i)
and frictional force at the wall,
Fw = μw Rw = 0.4 Rw ...(ii)
Resolving the forces vertically,
Rf + Fw = W + 0.5W = 1.5 W ...(iii)
and now resolving the forces horizontally,
Rw = Ff = 0.5 Rf or Rf = 2Rw
Now substituting the values of Rf and Fw in equation (iii),
2 Rw + 0.4 Rw = 1.5 W
∴
1.5
0.625
w 2.4
W
R ==W
and Fw = 0.4 Rw = 0.4 × 0.625 W = 0.25 W ...(iv)
Taking moments about A and equating the same,
(W × 2 cos 45°) + (0.5 W × x cos 45°)
= (Rw × 4 sin 45°) + (Fw × 4 cos 45°)
Substituting values of Rw and Fw from equations (iii) and (iv) in the above equation,
(W × 2 cos 45°) + (0.5 W × x cos 45°)
= (0.625 W × 4 sin 45°) + (0.25 W × 4 cos 45°)
Dividing both sides by (W sin 45°)*,

2 + 0.5 x = 2.5 + 1 = 3.5

∴ 3.5 – 2 3.0 m

0.5

x==^ Ans.

Fig. 9.4.