Why does the solution divide the weights by 2? (Highlighted yellow) 11-18.A 5-kg uniform serving table is supported on each side bytwo pairs of identical links, AB and CD, and springs CE. Ifthe bowl has a mass of 1 kg and is in equilibrium when0 = 45°, determine the stiffness k of each spring. The springsare unstretched when 0 = 90°. Neglect the mass of the links.SOLUTIONFree-Body Diagram: When 0 undergoes a positive virtual angular displacement of80, the dash line configuration shown in Fig. a is formed. We observe that only thespring force Fp, the weight W, of the table, and the weight W, of the bowl do workwhen the virtual displacement takes place. The magnitude of Fp can be computedusing 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 arespecified 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 080YG₁ = 0.25 sin 0 + bYG, = 0.25 sin 0 +xc= 0.25 cos(1)(2)Syg, = 0.25 cos 0808xc = -0.25 sin 080(3)Virtual Work Equation: Since W₁, W₁, and Fsp act towards the negative sense oftheir corresponding virtual displacement, their work is negative. Thus,SU = 0;WÁG + (−W‚Ôyg,) + (−Fspôxc) = 0- (-1)09.8Substituting 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) = 080-7.3575 cos 0 + 0.0625k sin cos 0) = 0Since 800, then-7.3575 cos 0 + 0.0625k sin cos 0 = 0117.72k =sinWhen = 45°, thenk =117.72sin 45°(9.81) = 24.525 N,Gb= 166 N/m(981) (981) N. Уст0.255inom 0.25sin@m(a)(4)Ans.-Esp250 mm150 mm.250 mm• DatumCBk150 mmSEDAns:k = 166 N/m

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Mechanical Engineering

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

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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