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 F⃗ required to slide the block C with constant speed is ____ N . (Use g=10 m/s² )

Solution

Physics sir · 4 steps

Draw B's free-body diagram first to get the tension, then add up everything that pulls back on C.

1. Read the setup. A is tied to the wall, so it stays still. B and C are joined by one string over a pulley fixed to the wall. When C moves left at constant speed, B moves right at the same speed, and the string has the same tension T on both sides.
2. Forces on B. B slides right under A, which presses on it with 40 N. That gives a friction of 0.5 × 40 = 20 N to the left. B also slides right over C, which pushes up on it with (4+6) × 10 = 100 N. That gives 0.5 × 100 = 50 N to the left. At constant speed the net force on B is zero, so T = 20 + 50 = 70 N.
3. Forces on C. C moves left, so every resisting force on it points right. The string pulls it with T = 70 N. B exerts the reaction friction of 50 N. The ground presses up with (4+6+8) × 10 = 180 N, so ground friction is 0.5 × 180 = 90 N.

   F = T + f_(BC) + f_(ground) = 70 + 50 + 90

4. So. The force needed to slide C at constant speed is 210 N.

The common slip: 190. Leaving out the friction between A and B makes T = 50 N and gives 190, not 210.

The idea to keep. When blocks are joined by a string over a pulley, first solve the block whose forces you fully know to get T. Every contact surface adds friction, and its normal force is the weight of everything above it.

Answer from the NTA final answer key. Solution by PrepWiser.