Ok, I understand that the equatio for a moment isM=rFsin. So how come when addin

Question

Ok, I understand that the equatio for a moment isM=rFsin. So how come when adding the moments together for force 3 thereis a cosine involved in the solution? I tried this problem withM=-2(300)sin60+5(400)sin120-(42+52)F3sin53.13. I am not getting the right answer, but I don't understandwhy. Ok, I understand that the equatio for a moment isM=rFsin. So how come when adding the moments together for force 3 thereis a cosine involved in the solution? I tried this problem withM=-2(300)sin60+5(400)sin120-(42+52)F3sin53.13. I am not getting the right answer, but I don't understandwhy.

Explanation / Answer

The equation for the Moment about A is actually... M = -300 cos 30o (2m) + 300 sin30o (0m) + 400 sin 60o (5m) - 400 cos60o (0m) + F3 cos (4m) - F3sin (5m) =       -4800 N. m When you are taking moments you have to take into account magnitudeand direction of all forces, in this case since the first twoforces are on the same plane as the moment, then the distance inthe y direction is 0 therefore eliminating those components of theequation.

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