Determine thetensions in cables AB and AC for values of d = 0, d = 0.9, d =1.8.

Question

Determine thetensions in cables AB and AC for values of d = 0, d = 0.9, d =1.8. (Note: Assume thatpoint A is fixed, and the length of AC will vary as distance dchanges.) PICTURELOCATION: https://s3.amazonaws.com/answer-board-image/201061232316341103015130512501216.jpg Determine thetensions in cables AB and AC for values of d = 0, d = 0.9, d =1.8. (Note: Assume thatpoint A is fixed, and the length of AC will vary as distance dchanges.) PICTURELOCATION: https://s3.amazonaws.com/answer-board-image/201061232316341103015130512501216.jpg

Explanation / Answer

Let the angle between the upper wall and AB be , angle between side wall and AC be . The forces are assumed to be directed towards the wall atA(tension) Then tan =1/1 =1 =>=45degree always tan =1/(1-d) for d=45deg Then for equilibrium of point A , balancing horizontal forces TAB cos = TAC sin since sin =cos =1/2 TAB =TAC Also for vertical forces TAB sin + TAC cos =(50*9.8)N=490N so TAB =TAC = 346.48 N b)balancing horizontal forces TAB cos = TAC sin =>TAB = TAC *.995 Also for vertical forces TAB sin + TAC cos =(50*9.8)N=490N so solving TAC = 447.7 TAB =445.45 c) =tan-1 1/(1.8-1) =51.34deg seeing F.B.D TAC sin= TAB cos and TAC cos +490 =TABsin solving , TAB =3463.9N TAC =3118.4N

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