Determine tensions in cables AC and BC given α = 20° and 12 kN load.
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In this engineering mechanics problem, we have a crane with a suspended load of 12 kilo-newtons. Point A is at the top of the crane boom, point C is where the cables meet the load, and point B is an anchor on the ground. We need to find the tensions in cables AC and BC.
To solve for the tensions, we'll draw a free-body diagram at point C. The forces acting on point C are: the tension in cable AC, the tension in rope BC, and the downward load of 12 kilo-newtons. We'll resolve these forces into their components and apply equilibrium equations.
We'll now resolve the forces into x and y components. For T_AC, the angle from vertical is 5 degrees, so its components are T_AC times sine of 5 degrees in the x-direction and T_AC times cosine of 5 degrees in the y-direction. For T_BC, the angle from horizontal is alpha equals 20 degrees, so its components are T_BC times cosine of 20 degrees in the x-direction and negative T_BC times sine of 20 degrees in the y-direction.
For equilibrium, the sum of forces in x and y directions must be zero. In the x-direction: T_AC times sine of 5 degrees minus T_BC times cosine of 20 degrees equals zero. In the y-direction: T_AC times cosine of 5 degrees plus T_BC times sine of 20 degrees minus 12 equals zero. We now have two equations with two unknowns, T_AC and T_BC, which we can solve simultaneously.
From the x-direction equation, we can express T_AC in terms of T_BC: T_AC equals T_BC times cosine of 20 degrees divided by sine of 5 degrees. Substituting this into the y-direction equation gives us: T_BC times cosine of 20 degrees divided by sine of 5 degrees, all times cosine of 5 degrees, plus T_BC times sine of 20 degrees equals 12. Solving for T_BC, we get T_BC equals 12 divided by cosine of 20 degrees times cotangent of 5 degrees plus sine of 20 degrees, which is approximately 1.15 kilo-newtons. Then T_AC equals T_BC times cosine of 20 degrees divided by sine of 5 degrees, which is approximately 12 kilo-newtons.
After solving the equilibrium equations, we find that the tension in cable AC is 12 kilo-newtons and the tension in rope BC is 1.15 kilo-newtons. This matches the second option provided in the question.