Question

The two uniform bars are identical. The mass and length of each bar are m= 30 kg and L=1.6 m. The top bar rotates around point A, coordinate , locates the bar, and force F(t) = 500 sin 60t (in N) is being applied to the end of the bar. A spring with k = 10000 N/m connects the top bar to a fixed point. The bottom bar rotates around its center at point B and coordinate 0g locates this bar. The bottom bar and two springs, each having spring constant k,, are being used as a vibration absorber for the top bar. a) Draw a free body diagram for each bar and derive the equations of motion. b) Find the value of k, needed to make the bottom bar act as a vibration absorber. c) Find the amplitude of the vibration of the bottom bar. 1.6 m F(t) = 500 sin 601 0.8 m A k, = 10000 N/m k, = ? B தே k, = ?

Answer 1

vens 200F undamped face system Thalysis Soonyoung 1.6 → -0.8m OD 104 ka ✓ OAL Ryp? soos in Cat TOA R2 xlsis@nx 0.8.- bilCine-1-6) & ois sinds) On RE? boxO-6. Comodoret ks Ana me JB=no mea: 256 Toba 12 = 6.4 122x0 ssinos FBD of syste 2 fro. ( sio 20) 25 Wurg Newton law aras Ten + 0.8C PA+ARU bis.cbxift+27 TO + ko'p, Liron + k240.8"Cümln - Neg) - FO). 0.8°CR7 -tax101) on – D. ŠP2O3 = (500 sin but )x16-00 For TB Trön + R2xo.s(sisob-ühOn) + 0.82 Resis on = 0 FONS 6-400 + 0.g? (R2+b)) OR -0.8°R2 04 2

Pore the hor B to aut as hesalios absorbere, natural fruan of hare na natural felen of bar B 슈 WA 12 WA KB JB 09 번 2 1.63k 40.8 kz 2R2x0.82 2.6 6:4 za roof bonos 12,9-1 =24 we get 0.82xt. Now k=101 N/m R2=5.71x103 N/m Now anpeiterele ole übration of boden bar is folk2 niet X2 500 Xl. 6 2x 5.71 X105x0.82 X2 = 0.109 m.

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