Engr Help

Example - Force –System Resultants -- Moments (continued)

Determine coordinates: A (0,2.1,0), B(0,1.4,0), C(0,0.6,0), D(0,0,0), E(0,-0.5,0)

F(0,-0.5,1.0), G(-2.4,0,0), H(0.4,0,-1.2) Determine force vectors:

F₁ = F_1x i + F_1y j + F_1z k , F₂ = F_2x i + F_2y j + F_2z k , F₃ = F_3x i + F_3y j + F_3z k

So the resultant force is

F_r = F₁ + F₂ + F₃ = ( F_1x + F_2x + F_3x ) i + (F_1y + F_2y + F_3y ) j + ( F_1z + F_2z + F_3z ) k

F₁ = 0 i + 0 j - 500 k, F₂ = 300 u_CH, F₃ = 700 u_FG

where u_CH is the unit vector from C to H, u_FG is the unit vector from F to G

CH = < 0.4,0,-1.2> - <0,0.6,0> = 0.4i – 0.6j -1.2k and |CH| = √(0.4² + 0.6² + 1.2²)

FG = < -2.4,0,0> - <0,-0.5.1.0> = -2.4i + 0.5j -1.0k and |FG| = √(2.4² + 0.5² + 1²)

u_CH = CH / |CH| = 0.286 i – 0.429 j – 0.867 k, F₂ = 300 [0.286 i – 0.429 j – 0.867 k]

u_FG = FG / |FG| = -0.906 i + 0.189 j – 0.378 k, F₃ = 700 [-0.906 i + 0.189 j – 0.378 k]

F₂ = 85.7 i -128.6 j -257.1 k, F₃ = -634.5 i +132.2 j -264.4 k

So the resultant force is F_r = F₁ + F₂ + F₃ = -548 i + 3.6 j – 1021 k N (result)

Now calculate the moment of each force vector about point A. The sum of these moments

equals the resultant moment, M_rA . You need the position vectors from A to any convenient

point on the line of action of the three force vectors.

for F₁ , r_AB = -0.7 j, for F₂, r_AC = -1.5 j , for F₃ , r_AG = -2.4 i -2.1 j

M_rA = r_AB x F₁ + r_AC x F₂ + r_AG x F₃ = 1290 i -634 j – 1521 k Nm (result)