A rectangular plate OBCD is 4 m long (OB, along the x-axis) and 3 m high (OD, along the y-axis). Determine the magnitude and direction of the resultant of the force system acting on it, and locate the point where the resultant cuts the x-axis (line O…

Answer: \(R=170.3\) N at \(\theta=40.24^\circ\) above the +x direction (up and to the right); it cuts the x-axis at 4.55 m to the right of O.

Take O as origin, x to the right, y upward, anticlockwise moments positive. Resolve the inclined forces: DB has slope \(3:4\) (length 5), so the 250 N force has components \((+200,-150)\) N; OC has slope \(3:4\), so the 100 N force has components \((+80,+60)\) N.

Sum of components:

\(\sum F_x=-150+200+80=130\) N; \(\sum F_y=200-150+60=110\) N.

\[R=\sqrt{130^2+110^2}=170.3\ \text{N},\qquad\theta=\tan^{-1}\frac{110}{130}=40.24^\circ\]

Both components are positive, so R acts up and to the right (first quadrant).

Moments about O (\(M=xF_y-yF_x\)):

• 200 N at B(4, 0), upward: \(+4\times200=+800\) N m
• 150 N at C(4, 3), leftward: \(-3\times(-150)=+450\) N m
• 250 N at D(0, 3), components (200, −150): \(0-3\times200=-600\) N m
• 100 N at O: passes through O, moment 0
• Couple: \(-150\) N m (clockwise)

\(\sum M_O=800+450-600+0-150=+500\) N m (anticlockwise).

Position of R (Varignon's theorem): if R cuts the x-axis at distance \(x\) from O, only its y-component has a moment about O:

\[x\cdot\sum F_y=\sum M_O\Rightarrow x=\frac{500}{110}=4.55\ \text{m to the right of O}\]

(Perpendicular distance of R from O \(=500/170.3=2.94\) m.)