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Why does the solution divide the weights by 2? (Highlighted yellow)

Why does the solution divide the weights by 2? (Highlighted yellow)

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Why does the solution divide the weights by 2? (Highlighted yellow)

11-18. A 5-kg uniform serving table is supported on each side by two pairs of identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg and is in equilibrium when 0 = 45°, determine the stiffness k of each spring. The springs are unstretched when 0 = 90°. Neglect the mass of the links. SOLUTION Free-Body Diagram: When 0 undergoes a positive virtual angular displacement of 80, the dash line configuration shown in Fig. a is formed. We observe that only the spring force Fp, the weight W, of the table, and the weight W, of the bowl do work when the virtual displacement takes place. The magnitude of Fp can be computed using the spring force formula, Fsp = kx = k(0.25 cos 0) = 0.25 k cos 0. Virtual Displacement: The position of points of application of W₁, W₁, and Esp are specified by the position coordinates yG YG, and x, respectively. Here, yG and YG, are measured from the fixed point B while xc is measured from the fixed point D. SyG= 0.25 cos 080 YG₁ = 0.25 sin 0 + b YG, = 0.25 sin 0 + xc= 0.25 cos (1) (2) Syg, = 0.25 cos 080 8xc = -0.25 sin 080 (3) Virtual Work Equation: Since W₁, W₁, and Fsp act towards the negative sense of their corresponding virtual displacement, their work is negative. Thus, SU = 0; WÁG + (−W‚Ôyg,) + (−Fspôxc) = 0 - (-1)09.8 Substituting W, = (9.81) = 4.905 N, W, = Fsp = 0.25k cos 0 N, Eqs. (1), (2), and (3) into Eq. (4), we have -4.905(0.25 cos 080) - 24.525(0.25 cos 080) - 0.25k cos 0(-0.25 sin 080) = 0 80-7.3575 cos 0 + 0.0625k sin cos 0) = 0 Since 800, then -7.3575 cos 0 + 0.0625k sin cos 0 = 0 117.72 k = sin When = 45°, then k = 117.72 sin 45° (9.81) = 24.525 N, Gb = 166 N/m (981) (981) N . Уст 0.255inom 0.25sin@m (a) (4) Ans. -Esp 250 mm 150 mm. 250 mm • Datum C B k 150 mm SE D Ans: k = 166 N/m
Transcribed Image Text:11-18. A 5-kg uniform serving table is supported on each side by two pairs of identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg and is in equilibrium when 0 = 45°, determine the stiffness k of each spring. The springs are unstretched when 0 = 90°. Neglect the mass of the links. SOLUTION Free-Body Diagram: When 0 undergoes a positive virtual angular displacement of 80, the dash line configuration shown in Fig. a is formed. We observe that only the spring force Fp, the weight W, of the table, and the weight W, of the bowl do work when the virtual displacement takes place. The magnitude of Fp can be computed using the spring force formula, Fsp = kx = k(0.25 cos 0) = 0.25 k cos 0. Virtual Displacement: The position of points of application of W₁, W₁, and Esp are specified by the position coordinates yG YG, and x, respectively. Here, yG and YG, are measured from the fixed point B while xc is measured from the fixed point D. SyG= 0.25 cos 080 YG₁ = 0.25 sin 0 + b YG, = 0.25 sin 0 + xc= 0.25 cos (1) (2) Syg, = 0.25 cos 080 8xc = -0.25 sin 080 (3) Virtual Work Equation: Since W₁, W₁, and Fsp act towards the negative sense of their corresponding virtual displacement, their work is negative. Thus, SU = 0; WÁG + (−W‚Ôyg,) + (−Fspôxc) = 0 - (-1)09.8 Substituting W, = (9.81) = 4.905 N, W, = Fsp = 0.25k cos 0 N, Eqs. (1), (2), and (3) into Eq. (4), we have -4.905(0.25 cos 080) - 24.525(0.25 cos 080) - 0.25k cos 0(-0.25 sin 080) = 0 80-7.3575 cos 0 + 0.0625k sin cos 0) = 0 Since 800, then -7.3575 cos 0 + 0.0625k sin cos 0 = 0 117.72 k = sin When = 45°, then k = 117.72 sin 45° (9.81) = 24.525 N, Gb = 166 N/m (981) (981) N . Уст 0.255inom 0.25sin@m (a) (4) Ans. -Esp 250 mm 150 mm. 250 mm • Datum C B k 150 mm SE D Ans: k = 166 N/m
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