Answers
To find the tension on the cable we have to use the total moments on the system about the pivot point is zero at equilibrium.
Moment = force × perpendicular distance.
L. Dism twing now at equirium, the tothe moments asout due prot will no nero, mro ut the tengor on the love ne T.
If we coryder the cookwise direction an are, mon the weight of the bridge acts on the costre of mass, moment due to the weight of the bridge, mir Fw Che) eso angl coso mi s 44 690 NX (10.9 m) eos 2,3,5° s 132699.6249 N. moment due to the caste, Ma-To A Siro) --TX 3.80 sm 33,5' mimo -TX 3.80 Sim 3,5 4 192627.6249 20 T2/ 192627 16299 - 31893 N. 3.80x sinBB5 (A) Tenager in the cable & 318930. now the component of force due to hing a densson of the case FT3184 y componert of force due to hing weight of the bridge. for tw = 44000 N .
so, que magretude of fore, en tant l 2. so, the force 101838N 101838N a templates 256 due to the hing has magridade of ( To force will act ach an argle 25.60 asore horizontal. 0 of the carre brears, the for one due to the odge at the about kivot point, n. amg) wsco a FW ( ) wsa 5 4400 NX ( 10,m) W313,5 =192127.1949 m. now duo moment of mentre of the bridge about The Pirot point, se emitex (A x (10.6m? 165007 am Rus 10x todnosn - rus утас, ar.
.
192627.6299 N 16500 un 1117 radlst of the bridge 12 so, precas arguar acceleration is 1.17 reals.
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