Question 3.22: A 500 N cylinder, 1 m in diameter is loaded between the cros...
A 500 N cylinder, 1 m in diameter is loaded between the cross pieces AE and BD which makes an angle of 60° with each other and are pinned at C. Determine the tension in the horizontal rope DE, assuming that the cross pieces rest on a smooth floor.
(UPTU I Sem, 2001–02)
Let the reaction at A and B is N_{A} and N_{B}, respectively
Thus, if total setup is considered for equilibrium
Then, ΣY = 0,
N_{A} + N_{B} = 500 N
As the setup is symmetrical thus N_{A} = N_{B}.
∴ N_{A} = N_{B} = 250 N
Considering FBD of cylinder as shown in Fig. 3.33 (a)
ΣY = 0, F_{1} \cos 60^{\circ}+F_{2} \cos 60^{\circ}=500
ΣX = 0, F_{1} \sin 60^{\circ}=F_{2} \sin 60^{\circ}
i.e., F_{1}=F_{2}
2 F_{1} \cos 60^{\circ}=500
F_{1}=F_{2} = 500 N
\sum M_{C} = 0, consider FBD of AE of Fig. 3.33 (a),
\Delta O L C, \tan 60^{\circ}=\frac{C L}{05}, C L=0.5 \tan 60^{\circ}
N_{A} \times 1.2 \cos 60^{\circ}+F_{1} \times C L=T \times C E \sin 60^{\circ}
N_{A} \times 1.2 \cos 60^{\circ}+F_{1} \times 0.5 \tan 60^{\circ}=T \times 1.8 \sin 60^{\circ}
250 \times 1.2 \cos 60^{\circ}+500 \times 0.5+60^{\circ}=T \times 1.8 \sin 60^{\circ}
T = 374 N
