Engineering Mechanics

(^154) „„„„„ A Textbook of Engineering Mechanics
Taking moments about A and equating the same,
(200 × 1.5 cos 60°) + 1000 × 3 cos 60°)
= (Fw × 3 cos 60°) + (Rw × 3 sin 60°)
Dividing both sides by the cos 60°,
300 + 3000 = (3 × Fw) + (3 × Rw tan 60°)
∴ 1100 = Fw + Rw tan 60° ...(iii)
We know that Fw = μw × Rw = 0.25 Rw ...(Qμw = 0.25)
Substituting this value of Fw in equation (iii),
1100 = (0.25 Rw) + (Rw tan 60°) = Rw (0.25 + 1.732) = Rw × 1.982
∴^1100 555 N
w 1.982
R ==
and Fw = 0.25 Rw = 0.25 × 555 = 138.7 N
Now substituting the value of Fw in equation (ii),
Rf + 138.7 = 1200
∴ Rf = 1200 – 138.7 = 1061.3 N
and Ff = μfRf = 0.35 × 1061.3 = 371.5 N
Now substituting the value of Ff in equation (i),
P + 371.5 = 555
∴ P = 555 – 371.5 = 183.5 N Ans.
(ii) Inclination of the ladder with the horizontal for no slipping
Now consider the ladder inclined at angle (α) and without any horizontal force acting at the
floor as shown in Fig. 9.6 (b).
Resolving the forces horizontally,
Rw = Ff = μf × Rf = 0.35 × Rf ...(iv)
and now resolving the forces vertically,
Rf + Fw = 1000 + 200 = 1200 N
We know that Fw = μw × Rw = 0.25(0.35 Rf) = 0.09 Rf ...(Q Rw = 0.35 Rf)
or Rf + 0.09 Rf = 1200
∴
1200
1101N
f 1.09
R ==
and Rw = 0.35Rf = 0.35 × 1101 = 385.4 N
Similarly Fw = 0.09 Rf = 0.09 × 1101 = 99.1 N
Taking moments about A and equating the same,
(1000 × 3 cos α) + (200 × 1.5 cos α)
= (Fw × 3 cos α) + (Rw × 3 sin α)