Question

Problem 5: A 5 kg uniform serving table and 1 kg bowl is supported by a series of links and springs. The near side of the table is supported by links AB and CD, and spring CE (as depicted). The far side of the table is supported by an identical pair of links and a spring (not shown). Using the principle of virtual work, determine the angle o where the table is in equilibrium. The springs each have a stiffness of k = 200 N/m and are unstretched when A= 90°. Neglect the mass of the links. 250 mm 150 mm o wwwww E ok 250 mm 150 mm

Answer 1

ANS! 2.50 150 mm 250mm Sybit Gb Wo / Si fub Neil Di G7X ette? Sac nc ESP A Ib you Bu D from the free body diagram, you can see o undergoes a positive visitual angular displacement of so, When virtual displacement takes place Spring force Esp weight of Wt tabel 7 األد 11 من height of baul his Isp = kx = 200 (0.25 Coso)- 50 COSON The position of points of application of Wb , Wf and Esp are specified by Dosition co-ordinates YG br y Gy and xe, Respectively . Here YG and you are Measured from fixed point B while Xc is measured from fixed point Dr.

You = 0.25 Sing+b Sg y = 0.25 casoso – @ Y6y=0.25 Sinota 8964 =0.250058 80 ☺ Xc=0.25 COSB Sac =-0.25 Sin OSO - Since Ho, Wc , Isp act towards the negative sense of their corresponding displacement their work is negative. - Suso; - No By Gb +GW+ SYG+)+ (- Esp Sx c ) 20 - Substituting Wbs **9.81 =4.905 H wf = 5x9.81 = 24.526 N Fsp = 50 COSO N D - Equato, Ginto & -4.905 (0.25 cose so )–24.525C0-25 Cosofo), a 50 Case (-0.25 Sino so so I so (-7.3575 Cos@ +12-55100 COS 8)0 Since so to, then -7.3575 cos@ +12-5510 Cos 0-0 cos@ C-7-3575 +12-5 Sino) -o By solving COSO 20 0=900 -7.3575 +12.58ing -o 0 = 36.9