Static Equilibrium Question
Three cables are connected at a central point B. The first cable runs from point A to point B at an angle 50˚ above the horizontal. Another cable runs from point B to point C. This cable has a horizontal dimension of 12 and a vertical dimension of 5 (creating a pythagorean triple). The final cable runs straight down from point B and is connected to a box, D, that wieghs 500lbs.
Find the tension in each cable.
My Answer(s) so far:
Tension in cable TBD: 500lbs
Tension in cable TAB: -698.846lbs (which doesn't make sense)
Tension in cable TBC: 1003.580lbs
I found the tension in cable TBD to be the weight of the box since the ∑F (in the y direction) at box D is zero:
∑F = 0, therefore Fy = W = 500lbs
To find the tension of cables TAB and TBC I found up the following system of equations
TBC*(5/13) + TABsin(50˚) = 500
TBC*(12/13) + TABcos(50˚) = 0
The answer is supposed to be 500lbs for TBD, 483.62lbs for TAB, and 310.86 for TBC
My mistake is somewhere in the system of equations but I am not sure what part is wrong. Thanks!