The bent rod in Fig. a is supported at A by a journal bearing, at D by a ball-and-socket joint, and at B by means of cable BC. Using only one equilibrium equation, obtain a direct solution for the tension in cable BC. The bearing at A is capable of exerting force components only in the z and y directions since it is properly aligned on the shaft. In other words, no couple moments are required at this support.

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
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The bent rod in Fig. a is supported at A by a journal bearing, at D by
a ball-and-socket joint, and at B by means of cable BC. Using only
one equilibrium equation, obtain a direct solution for the tension in
cable BC. The bearing at A is capable of exerting force components
only in the z and y directions since it is properly aligned on the shaft.
In other words, no couple moments are required at this support.
1m
B.
0.5 m
Free-Body Diagram. As shown in Fig. b, there are six unknowns
05 m
Equations of Equilibrium. The cable tension Tg may be obtained
directly by summing moments about an axis that passes through
points D and A. Why?
00 kg
(a)
Since the moment arms from the axis to T, and W are easy to obtain,
we can determine this result using a scalar analysis. As shown,Fig. b
EMDA = 0; Tạ(1 m sin 45°) – 981 N(0.5 m sin 45°) = 0
T = 490.5 N
0.5 m
W 981 N
05 m
D,
(b)
Transcribed Image Text:The bent rod in Fig. a is supported at A by a journal bearing, at D by a ball-and-socket joint, and at B by means of cable BC. Using only one equilibrium equation, obtain a direct solution for the tension in cable BC. The bearing at A is capable of exerting force components only in the z and y directions since it is properly aligned on the shaft. In other words, no couple moments are required at this support. 1m B. 0.5 m Free-Body Diagram. As shown in Fig. b, there are six unknowns 05 m Equations of Equilibrium. The cable tension Tg may be obtained directly by summing moments about an axis that passes through points D and A. Why? 00 kg (a) Since the moment arms from the axis to T, and W are easy to obtain, we can determine this result using a scalar analysis. As shown,Fig. b EMDA = 0; Tạ(1 m sin 45°) – 981 N(0.5 m sin 45°) = 0 T = 490.5 N 0.5 m W 981 N 05 m D, (b)
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