A force F acts at the top of end A of the pole. Its magnitude is 6kN and its x c
ID: 1820718 • Letter: A
Question
A force F acts at the top of end A of the pole. Its magnitude is 6kN and its x component is 4kN. The coordinates of point A are as shown. Determine the components of F so that the magnitude of the moment due to F about point P is as large as possible.
A force F acts at the top of end A of the pole. Its magnitude is 6kN and its x component is 4kN. The coordinates of point A are as shown. Determine the components of F so that the magnitude of the moment due to F about point P is as large as possible.Explanation / Answer
The r vector is rpa = (4m-0m)i + (3m-0m)j +(-2m-0m)k = 4mi + 3mj -2mk The magnitude of the force is F = 4kNi + Fyj +Fzk |F|^2 = (4kN)^2 +Fy^@ +Fz^2 =6kN^2 Fy^2 +Fz^2 = 20kN^2 In order to get the maximum moment due F, it must be perpendicular to the r vector. So use the dot product between the force and r vector and set it equal to zero. rpa dot F =(4mi + 3mj -2mk)dot(4kNi +Fyj +Fzk)=0 16kN-m + (3m)Fy -(2m)Fz =0 Fz = 1.5Fy +8kN Two equations and two unknowns substitute the second equation into the first equation. Fy^2 +(1.5Fy +8Kn)^2 = 20kN^2 3.25Fy^2 + 24kNFy + 44kN^2 = 0 Fy= -24kN +(or -) sqrt(24kN^2 - 4(3.25)(44Kn^2) = -3.385kN or -4kN When Fy = -3.385 Fz = 2.923 When Fy = -4kN Fz= 2kN The force components are (4 kN, -3.385 kN, 2.923 kN) or (4 kN, - 4 kN, 2 kN).
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