(^156) A Textbook of Engineering Mechanics
Taking moments of the forces about the block A and equating the same,
Rw × l cos 45° + Fw × l cos 45° = W × l cos 45°
Rw + Fw = W
or Rw + μRw = W
Rw (1 + μ) = W
Substituting the value of Rw from equation (iv),
2
2
(1 )
1
W
W
μ×
+μ =
μ+
or 2 μ (1 + μ) = μ^2 + 1
∴ 2 μ + 2μ^2 = μ^2 + 1
μ^2 + 2μ – 1 = 0
Solving it as quadratic equation for μ.
–2 (2)^24
0.414
2
±+
μ= =^ Ans.
EXERCISE 9.1
- A 4 m ladder weighing 250 N is placed against a smooth vertical wall with its lower end
1.5 m away from the wall. If the coefficient of friction between the ladder and the floor is
0.3, show that the ladder will remain in equilibrium in this position. - A ladder shown in Fig. 9.9 is 4 m long and is supported by a horizontal floor and vertical
wall. The coefficient of friction at the wall is 0.25 and that at the floor is 0.5. The weight
of the ladder is 30 N and is considered to be concentrated at G. The ladder also supports
a vertical load of 150 at C.
Fig. 9.9.
Determine the reactions at A and B and compute the least value of (α) at which the ladder may
be placed without slipping to the left.
[Ans. RA = 178.9 N; RB = 82.46 N; α = 53.3°]