1. **State the problem:** We have a pulley system with two blocks, B and D, connected by a rope over pulleys. Block D has a mass of 5 kg and hangs vertically a distance $a=2.6$ m below pulley C. The horizontal distance between points A and C is $b=0.333$ m. We need to find the mass of block B such that the system is in equilibrium. 2. **Analyze the system:** For equilibrium, the tension in the rope must balance the weights of the blocks. The rope length is constant, so the vertical displacement of block B relates to the horizontal distance $b$ and vertical distance $a$. 3. **Use geometry:** The rope from A to B to C forms a right triangle with horizontal side $b$ and vertical side $a$. The length of rope segment from A to B to C is $$L = \sqrt{a^2 + b^2}$$ 4. **Calculate $L$:** $$L = \sqrt{(2.6)^2 + (0.333)^2} = \sqrt{6.76 + 0.111} = \sqrt{6.871} \approx 2.62$$ 5. **Relate masses for equilibrium:** The tension $T$ in the rope supports both blocks. The weight of block D is $$W_D = m_D g = 5 \times 9.8 = 49 \text{ N}$$ 6. **Balance forces:** The tension must balance the weight of block B, but because of the pulley arrangement, the effective force on block B is related to the rope length and geometry. The vertical component of tension supporting block B is proportional to $a/L$. 7. **Calculate mass of block B:** Using the ratio of vertical distances, $$m_B g = W_D \times \frac{b}{a}$$ Since block B hangs vertically, the force balance is $$m_B = m_D \times \frac{b}{a} = 5 \times \frac{0.333}{2.6} = 5 \times 0.128 = 0.64$$ 8. **Final answer:** The mass of block B is approximately **0.64 kg**.