The force F 9i kN acts on the ring A where the cables AB AC

The force F = 9i (kN) acts on the ring A where the cables AB, AC, and AD are joined What is the sum of the moments about point D due to the force F and the three forces exerted on the ring by the cables? Strategy: The ring is in equilibrium Use what you know about the four forces acting on it

Solution

Position Vectors

OA= 12i+4j+2k

OD=6j

OC=4j+6k

OB=6i

Force F= 9i KN

vector along AC = OA – OC = (12i-2j+2k)

Unit vector along AC = (12i-2j+2k)/sqrt(12*12+2*2+2*2)

Unit vector along AC = (12i-2j+2k)/12.32

Similarly

Unit vector along AB=(6i+4j+2k)/7.48

Tensions in the Cable AC = component of force F along AC

Tac= 9i.unit vector along AC = 8.766233 KN

Tac vector = 8.766((12i-2j+2k)/12.32) = 8.53i -1.422j+1.422k

Tensions in the Cable AB = component of force F along AB

Tab= 9i.unit vector along AB = 7.2192 KN

Tab Vector =7.2192((6i+4j+2k)/7.48) =5.79i +3.86j +1.93k

Moment of force Tad along cable AD is zero because it passes through point D.

As per Varignon’s Theorem

Sum of Moment of concurrent forces about D is = OD x (vectorial sum of all forces at point A)

Moment about D = MD = 6j x ( 9i + 8.53i -1.422j+1.422k +5.79i +3.86j +1.93k)

MD= 6j x (23.32i +2.44j + 3.352k)

MD = -139.92 k +0+20.112i KNm