In the given figure the blocks A, B and C weigh 4 ~kg , 6 ~kg and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5. The force F required to slide the block C with constant speed is _ _ _ _ N. (Use g=10 ~m / s ² )

Step-by-step solution

For the 8 kg block (C) to move with constant velocity, F_ net = 0 . Friction forces acting: - Between C and ground: f = (m_A + m_B + m_C)g = 0.5 (4+6+8) 10 = 90 N - Between C and B: f = (m_A + m_B)g = 0.5 (4+6) 10 = 50 N For the 6 kg block (B): T = f_ B on ground + f_ A on B T = (m_A + m_B)g m_B m_A+m_B + m_A g T = 20 + 50 = 70 N For the 8 kg block (C): F = 90 + T + 50 = 90 + 70 + 50 = 210 N