Answer:
Fad = 28.8 kN
Fbd = 16.4 kN
Fcd = 28.1 kN
Explanation:
First, find the length of each cable.
AD = √((2 m)² + (0.5 m)² + (2.5 m)²)
AD = √10.5 m
AD ≈ 3.24 m
BD = √((1.5 m)² + (1 m)² + (2.5 m)²)
BD = √9.5 m
BD ≈ 3.08 m
CD = √((1 m)² + (1 m)² + (2.5 m)²)
CD = √8.25 m
CD ≈ 2.87 m
Next, use similar triangles to find the x, y, and z components of each tension force.
Fadx = 2/3.24 Fad = 0.617 Fad
Fady = 0.5/3.24 Fad = 0.154 Fad
Fadz = 2.5/3.24 Fad = 0.772 Fad
Fbdx = 1.5/3.08 Fbd = 0.487 Fbd
Fbdy = 1/3.08 Fbd = 0.324 Fbd
Fbdz = 2.5 / 3.08 Fbd = 0.811 Fbd
Fcdx = 1/2.87 Fcd = 0.348 Fcd
Fcdy = 1/2.87 Fcd = 0.348 Fcd
Fcdz = 2.5/2.87 Fcd = 0.870 Fcd
Now sum the forces in the x, y, and z directions:
∑Fx = ma
-0.617 Fad + 0.487 Fbd + 0.348 Fcd = 0
∑Fy = ma
-0.154 Fad − 0.324 Fbd + 0.348 Fcd = 0
∑Fz = ma
60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0
To solve this system of equations algebraically, start by subtracting the first two equations, eliminating Fcd.
-0.463 Fad + 0.811 Fbd = 0
0.811 Fbd = 0.463 Fad
Fbd = 0.571 Fad
Substitute into either of the first two equations:
-0.617 Fad + 0.487 (0.571 Fad) + 0.348 Fcd = 0
-0.617 Fad + 0.278 Fad + 0.348 Fcd = 0
-0.339 Fad + 0.348 Fcd = 0
0.348 Fcd = 0.339 Fad
Fcd = 0.975 Fad
Now substituting into the third equation:
60 kN − 0.772 Fad − 0.811 Fbd − 0.870 Fcd = 0
60 kN − 0.772 Fad − 0.811 (0.571 Fad) − 0.870 (0.975 Fad) = 0
60 kN − 0.772 Fad − 0.463 Fad − 0.849 Fad = 0
60 kN − 2.083 Fad = 0
Fad = 28.8 kN
Solving for the other two tension forces:
Fbd = 0.571 Fad = 16.4 kN
Fcd = 0.975 Fad = 28.1 kN
