I have a question about the direction of forces when drawing free body diagrams.
ID: 1393781 • Letter: I
Question
I have a question about the direction of forces when drawing free body diagrams.
In the picture provided, I drew the FBD exactly how it is except that I drew the tension in cable BC as point up, not down. I then proceeded to take the moment about point A in order to find the value of the tension in BC (which is 34.6). With the value of BC known, I then plugged that value into the equation of equilibrium in the y direction, which is: Fy = 0 = Ay - 60 + BC(4/5). Solving, for Ay, I got 32.32 lb.
However, if I solve the problem using the FBD the way it is shown in the picture (with BC point down, not up) and plug in 34.6 into Fy = 0 = Ay - 60 - BC(4/5), I get 87.6 lb as the value for Ay.
Why is this? I thought I was told that since we dont know the true direction of the forces when drawing the FBD, it doesnt matter if we draw the forces as up or down because, no matter in which direction we drew the forces, we would get the same value but with different signs.. Yet, here is obviously does matter because both values obtained are completely different. Also, how can I know, for the next time, the correct direction that for which I should draw tensions in cables? Is it that, when drawing the FBD, the force of tension in cables should always point away from the strucutre?
Explanation / Answer
In order for the system to be in equilibrium, not just the moment should be zero but also the net forces
should be zero. If you assume that the tension in the cable BC is pointing up, then the net force
in the x axis will be different from zero.
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