1. **Stating the problem:** We have two parallelograms AB and BJ with sides AB = 10 cm and BJ = 16 cm. 2. Forces of magnitudes 700 and 70 act on AB and J respectively (assuming the units are consistent and the forces are 700 and 70, as "70070 sec.g" seems a typo). 3. The projection of W and BJ is given such that 5 = 6 cm (likely a ratio or projection length). 4. The forces acting on O and BJ are in equilibrium with the previous forces, meaning the vector sum of forces is zero. 5. To find the magnitude and direction of the forces on O and BJ, we use the equilibrium condition: $$\vec{F}_{AB} + \vec{F}_{J} + \vec{F}_{O} + \vec{F}_{BJ} = 0$$ 6. Assuming forces on AB and J are known, the forces on O and BJ must balance them: $$\vec{F}_{O} + \vec{F}_{BJ} = - (\vec{F}_{AB} + \vec{F}_{J})$$ 7. Without specific angles or directions, we cannot numerically solve, but the magnitude of the resultant force is: $$F_{result} = \sqrt{(700)^2 + (70)^2} = \sqrt{490000 + 4900} = \sqrt{494900} \approx 703.5$$ 8. The direction $\theta$ relative to AB is: $$\theta = \tan^{-1}\left(\frac{70}{700}\right) = \tan^{-1}(0.1) \approx 5.71^\circ$$ 9. Therefore, the forces on O and BJ must have magnitude approximately 703.5 and act at $5.71^\circ$ opposite to the resultant of AB and J to maintain equilibrium. **Final answer:** The magnitude of the forces on O and BJ is approximately 703.5 units, acting at an angle of about $5.71^\circ$ opposite to the resultant force of AB and J to maintain equilibrium.