Resultant magnitude, direction and position of forces

Reading of the figure: rectangle ABCD with A(0,0), B(5,0), C(5,3), D(0,3) (m). Forces:

• 200 kN at D, directed up and to the left at 30° to the horizontal.
• 300 kN at C, horizontal, to the right.
• 150 kN at B, directed up and to the right at 30° to the horizontal.
• 100 kN at A, horizontal, to the left.

Resultant force

ΣFx = −200 cos30° + 300 + 150 cos30° − 100
    = −173.21 + 300 + 129.90 − 100 = +156.70 kN
ΣFy = +200 sin30° + 150 sin30° = 100 + 75 = +175.00 kN

R = √(156.70² + 175² ) = 234.9 kN θ = tan⁻¹(175/156.70) = 48.16° above the horizontal (towards the right).

Moment about A (counter-clockwise positive, M = x·Fy − y·Fx)

200 kN at D(0,3):  0 − 3(−173.21) = +519.62
300 kN at C(5,3):  0 − 3(300)     = −900.00
150 kN at B(5,0):  5(75) − 0      = +375.00
100 kN at A:                      =    0
M_A                               =   −5.38 kN·m

Position

x · R_y − y · R_x = M_A

• On AB (y = 0): x = −5.38/175 = −0.031 m, that is 0.031 m to the left of A.
• On AD (x = 0): y = −M_A/R_x = 5.38/156.70 = 0.034 m above A.

So the resultant of 234.9 kN at 48.2° practically passes through A (it cuts AB 0.031 m to the left of A, and cuts AD 0.034 m above A).