A cylinder weighing 1000 N and 1.5 m diameter is supported by a beam AB of length 6 m and weight 400 N as shown in Fig. Neglecting Friction at the surface of contact of the cylinder, determine:
i) Wall reaction at D
ii) Tension in the cable BC
iii) Hinged Reaction at support A
Best answer
Answer: Wall reaction at D \(R_D = 1000\ \text{N}\) (horizontal), tension in cable BC \(T \approx 588.2\ \text{N}\), hinge reaction at A \(R_A \approx 1209.8\ \text{N}\) at \(66.1^\circ\) to the horizontal.
Data from the figure: beam AB = 6 m, hinged at A at the foot of the vertical wall and inclined at \(45^\circ\) to the horizontal; cable BC goes from B to point C on the wall and makes \(60^\circ\) with the wall; the cylinder (W = 1000 N, r = 0.75 m) rests between the wall (contact D) and the beam (contact P); beam weight 400 N acts at its midpoint.
1. Equilibrium of the cylinder
Forces: W = 1000 N (down), \(R_D\) (horizontal, from the wall), \(R_P\) (normal to the beam, i.e. at \(45^\circ\) to the vertical).
\(\sum F_y = 0:\ R_P \cos 45^\circ = 1000 \Rightarrow R_P = 1414.2\ \text{N}\)
\(\sum F_x = 0:\ R_D = R_P \sin 45^\circ = 1000\ \text{N}\)
2. Position of contact P on the beam
The wall and the beam meet at A at an angle of \(45^\circ\) (beam at \(45^\circ\) to horizontal, so also \(45^\circ\) to the vertical wall). The centre O lies on the bisector of this angle, so
\(AP = \dfrac{r}{\tan 22.5^\circ} = \dfrac{0.75}{0.4142} = 1.811\ \text{m}\)
3. Equilibrium of the beam AB
Forces on the beam: \(R_P = 1414.2\) N at P pressing on the beam (perpendicular to it), 400 N at the midpoint (3 m from A), T along BC at B, and the hinge reaction \((A_x, A_y)\) at A.
Coordinates of B: \(x_B = y_B = 6\cos 45^\circ = 4.243\ \text{m}\). The cable makes \(60^\circ\) with the vertical, so its components at B are \(T\sin 60^\circ\) towards the wall and \(T\cos 60^\circ\) upwards.
\(\sum M_A = 0:\)
\(T(\sin 60^\circ \times 4.243 + \cos 60^\circ \times 4.243) = R_P \times AP + 400 \times 3\cos 45^\circ\)
\(5.796\,T = 1414.2 \times 1.811 + 400 \times 2.121 = 2560.7 + 848.5 = 3409.2\)
\(T = 588.2\ \text{N}\)
4. Hinge reaction at A
Components of \(R_P\) on the beam: 1000 N horizontal (away from the wall) and 1000 N downward.
\(\sum F_x = 0:\ A_x + 1000 - T\sin 60^\circ = 0 \Rightarrow A_x = 509.4 - 1000 = -490.6\ \text{N}\) (i.e. 490.6 N towards the wall)
\(\sum F_y = 0:\ A_y - 1000 - 400 + T\cos 60^\circ = 0 \Rightarrow A_y = 1400 - 294.1 = 1105.9\ \text{N}\) (upward)
\(R_A = \sqrt{490.6^2 + 1105.9^2} = 1209.8\ \text{N}\), \(\ \alpha = \tan^{-1}\dfrac{1105.9}{490.6} = 66.1^\circ\) with the horizontal.
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