Explanation Assumption: The self-weight of the cable is ignored. The cable is perfectly flexible and inextensible. The tensile force acting in the cable is always tangent to the cable at points along its length. Being inextensible, the cable has a constant length both before and after the load is applied. The cable or a segment of the cable can be treated as a rigid body. Method of joints is used to determine the force in the cable. Consider the state of member as tension where the force is pulling the member and as compression where the force is pushing the member. Consider the force indicating the right side as positive and left side as negative in horizontal components of forces. Consider the force indicating upward is taken as positive and downward as negative in vertical components of forces. Consider clockwise moment as negative and anti-clock wise moment as positive wherever applicable. Apply Equation of equilibrium wherever applicable. Segment BCDE : Show the free body diagram of the segment BCDE as in Figure (1). Using Figure (1), Determine the value of sin θ : sin θ = Opposite side Hypotenuse (I) Determine the value of cos θ : cos θ = Adjacent side Hypotenuse (II) Moment about the point E : Determine the tension in the cable segment AB and the sag y B by taking moment about point E . ∑ M E = 0 200 ( 12 ) + 500 ( 27 ) + 300 ( 47 ) − T A B cos θ 1 ( y B + 4 ) − T A B sin θ 1 ( 47 ) = 0 (III) Conclusion: Substitute y B for opposite side and y B 2 + 144 for hypotenuse in Equation (I). sin θ 1 = y B y B 2 + 144 (IV) Substitute 12 ft for adjacent side and y B 2 + 144 for hypotenuse in Equation (II). cos θ 1 = 12 y B 2 + 144 (V) Substitute y B y B 2 + 144 for sin θ 1 and 12 y B 2 + 144 for cos θ 1 Equation (III). [ 2 , 400 + 13 , 500 + 14 , 100 − T A B 12 y B 2 + 144 ( y B + 4 ) − T A B y B y B 2 + 144 ( 47 ) ] = 0 30 , 000 − T A B ( 12 y B + 48 y B 2 + 144 ) − T A B ( 47 y B y B 2 + 144 ) = 0 30 , 000 − T A B ( 59 y B + 48 y B 2 + 144 ) = 0 T A B ( 59 y B + 48 y B 2 + 144 ) = 30 , 000 (VI) Segment BC : Show the free body diagram of the segment BC as in Figure (2). Using Figure (2), Moment about the point C : Determine the force in the cable segment AB and the sag y B by taking moment about point C . ∑ M C = 0 300 ( 20 ) + T A B cos θ 1 ( 14 − y B ) − T A B sin θ 1 ( 20 ) = 0 (VII) Conclusion: Substitute y B y B 2 + 144 for sin θ 1 and 12 y B 2 + 144 for cos θ 1 Equation (VII). 300 ( 20 ) + T A B 12 y B 2 + 144 ( 14 − y B ) − T A B y B y B 2 + 144 ( 20 ) = 0 6 , 000 + T A B ( 168 − 12 y B y B 2 + 9 ) − T A B ( 20 y B y B 2 + 9 ) = 0 6 , 000 − T A B ( 32 y B − 168 y B 2 + 144 ) = 0 T A B ( 32 y B − 168 y B 2 + 144 ) = 6 , 000 (VIII) Divide Equation (VI) by Equation (VIII). F A B ( 59 y B + 48 y B 2 + 144 ) F A B ( 32 y B − 168 y B 2 + 144 ) = 30 , 000 6 , 000 59 y B + 48 32 y B − 168 = 30 , 000 6 , 000 354 , 000 y B + 288 , 000 = 960 , 000 y B − 5 , 040 , 000 606 , 000 y B = 5 , 328 , 000 y B = 8.79 ft Substitute 8.79 ft for y B in Equation (VIII). T A B ( 32 ( 8.79 ) − 168 8.79 2 + 144 ) = 6 , 000 T A B ( 113.28 14.875 ) = 6 , 000 T A B = 787.87 lb Thus, the sag y B at point B of the cable is y B = 8.79 ft _ . Thus, the tension in the cable segment AB is T A B = 787.87 lb _ . Joint B : Show the free-body diagram of the joint B as in Figure (3). Using Figure (3), Determine the value of θ : tan θ = Opposite side Adjacent side (IX) Along the vertical direction: Determine the tension in the cable segment BC by resolving the vertical component of forces. ∑ F y = 0 T A B sin θ 1 − 300 − T B C sin θ 2 = 0 (X) Conclusion: Substitute 8