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FGH≔―――――――――――= ((-1.5 kN ⋅ 3 m -((-1 kN)) ⋅ 2.25 m)) 3 m -0.75 kN SITUATION 3. The towers at the opposite side of the river, supports the load W = 100 kN as shown in the figure. The distance L = 15 m and the sag d= 0.75 m. a) Tension in BC if x1 = 5m W≔100 kN θAC≔atan = ⎛ ⎜ ⎝―0.75―5 ⎞ ⎟ ⎠ 8.531 deg θBC≔atan = ⎛ ⎜ ⎝―0.75―15-5 ⎞ ⎟ ⎠ 4.289 deg ΣFx=0 FAC ⋅ cos ⎛⎝θAC⎞ ⎠=FBC ⋅ cos ⎛⎝θBC⎞ ⎠ FAC=FBC ⋅――― cos ⎛⎝θBC⎞ ⎠ cos ⎛⎝θAC⎞ ⎠ ΣFy=0 FAC ⋅ sin⎛⎝θAC⎞ ⎠+FBC ⋅ sin⎛⎝θBC⎞ ⎠-W=0 FBC ⋅―――⋅ + - = cos ⎛⎝θBC⎞ ⎠ cos ⎛⎝θAC⎞ ⎠ sin⎛⎝θAC⎞ ⎠ FBC ⋅ sin⎛⎝θBC⎞ ⎠ W 0 FBC ⎛⎝cos ⎛⎝θBC⎞ ⎠ ⋅tan⎛⎝θAC⎞ ⎠+sin⎛⎝θBC⎞ ⎠ ⎞ ⎠=W FBC≔――――――――――= W ⎛⎝cos ⎛⎝θBC⎞ ⎠ ⋅ tan⎛⎝θAC⎞ ⎠+sin⎛⎝θBC⎞ ⎠ ⎞ ⎠ 445.693 kN b) Total length if x1 = 5m

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