Question #96210
Questions in the link
https://eksiup.com/p/zh2525055pgx
1
Expert's answer
2019-10-10T09:43:35-0400

In general case, as the net force is zero:


TABcosθ2=Fsinθ1{T}_{AB}\cos\theta_2=F\sin\theta_1TAC+TABsinθ2=Fcosθ1T_{AC}+T_{AB}\sin\theta_2=F\cos\theta_1

It has the solution:


TAB=Fsinθ1cosθ2{T}_{AB}=\frac{F\sin\theta_1}{\cos\theta_2}


TAC=Fsinθ1cosθ2sinθ2+Fcosθ1T_{AC}=-\frac{F\sin\theta_1}{\cos\theta_2}\sin\theta_2+F\cos\theta_1

So, for case a):

TAB=100sin0cos45=0N{T}_{AB}=\frac{100*\sin0^{\circ}}{\cos45^{\circ}}=0N

TAC=0+100cos0=100NT_{AC}=-0+100\cos0^{\circ}=100N

for case b)



TAB=100sin30cos60=100N{T}_{AB}=\frac{100*\sin30^{\circ}}{\cos60^{\circ}}=100N

TAC=36.6NT_{AC}=36.6N


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