## Answers

Solution is attached in the image Posted below.

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Зк) ее SOK A Taking component of force. vertical 50 COSO 10 10 - 30 Horizontal - so sino you tanty 3 =) 53. 13° 30 3klft tid t no from hinge 3kift R Disconnect 30 ] Tuo uttet f 10 EMB=0 Rx10- 3XL0 ? = 0 2 R= ISK

3Kft 136 I LIS TTT с 'n is С 8 ५० LO 10 Ak, Ay, Am, Dy NOW taking CD: Efy=0 1st Dy= 3x10 30-15 2 - ISK Taking span AC Efy=0 Ay 3 Ay -30-15=0 [Ay= 45k & fr=0 Ac +40=0 AC = -uok [t MA= 15x20 + 30xio 600k-ft How overal Result AC = - Lok [2] Ay= usk [1] M4: 600 E-ft [2] Dy=isk [T]

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