Replace the loading by an equivalent resultant force and couple moment acting at point O. Assume F1 = {-420i + 210j + 280k} N and F2 = {-450k} N.

Steps to Solve

1. Calculate the resultant force The resultant force $F_R$ is the sum of the forces $F_1$ and $F_2$.

$$ F_R = F_1 + F_2 $$ $$ F_R = (-420i + 210j + 280k) + (0i + 0j - 450k) $$ $$ F_R = -420i + 210j - 170k \text{ N} $$

2. Calculate the moment due to $F_1$ about point O First find the position vector $r_{OB}$ from point O to point B. $$ r_{OB} = 2i + 1.5j \text{ m} $$ Then calculate the moment $M_{O1}$ due to $F_1$ about point O. $$ M_{O1} = r_{OB} \times F_1 = (2i + 1.5j) \times (-420i + 210j + 280k) $$ $$ M_{O1} = \begin{vmatrix} i & j & k \ 2 & 1.5 & 0 \ -420 & 210 & 280 \end{vmatrix} $$ $$ M_{O1} = (1.5 \cdot 280 - 0 \cdot 210)i - (2 \cdot 280 - 0 \cdot (-420))j + (2 \cdot 210 - 1.5 \cdot (-420))k $$ $$ M_{O1} = 420i - 560j + (420 + 630)k $$ $$ M_{O1} = 420i - 560j + 1050k \text{ N.m} $$

3. Calculate the moment due to $F_2$ about point O First find the position vector $r_{OA}$ from point O to point A. $$r_{OA} = 2i + 1.5j \text{ m}$$ Then calculate the moment $M_{O2}$ due to $F_2$ about point O. $$M_{O2} = r_{OA} \times F_2 = (2i + 1.5j) \times (0i + 0j - 450k)$$ $$M_{O2} = \begin{vmatrix} i & j & k \ 2 & 1.5 & 0 \ 0 & 0 & -450 \end{vmatrix}$$ $$M_{O2} = (1.5 \cdot (-450) - 0 \cdot 0)i - (2 \cdot (-450) - 0 \cdot 0)j + (2 \cdot 0 - 1.5 \cdot 0)k$$ $$M_{O2} = -675i + 900j + 0k \text{ N.m}$$

4. Calculate the resultant moment about point O The resultant moment $M_O$ is the sum of the moments $M_{O1}$ and $M_{O2}$. $$ M_O = M_{O1} + M_{O2} = (420i - 560j + 1050k) + (-675i + 900j + 0k) $$ $$ M_O = (420 - 675)i + (-560 + 900)j + (1050 + 0)k $$ $$ M_O = -255i + 340j + 1050k \text{ N.m} $$