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Acceleration of block A with blocks B and C

Short answer

Friction on A from B: 0.35×50 = 17.5 N; from table: 0.25×150 = 37.5 N; total 55 N. Dynamics: 500 – T = (500/9.81)a, T – 55 = (100/9.81)a. Adding gives a = (500−55)×9.81/600 = 7.28 m/s².

Determine the acceleration of block A for the system, if the system starts from rest. Coefficient of friction between block A and table is 0.25 and that between blocks A and B is 0.35. Weights of blocks A, B and C are 100 N, 50 N and 500 N respectively. [Figure: block B sits on block A on a horizontal table; B is tied to the wall by a rope on the left; A is connected over a pulley at the table edge to hanging block C.]

Reading of the figure

Block B rests on block A, both on a horizontal table. B is tied by a rope to the wall on the left, so it cannot move. A is connected by a rope over a pulley at the table edge to the hanging block C. When released, C falls, A moves to the right and slides under the fixed block B. Weights: A = 100 N, B = 50 N, C = 500 N. μ(table–A) = 0.25 and μ(A–B) = 0.35.

 wall |--rope--[B]
      |        [A]-----------o
                              |
                             [C]

Friction forces on A (A is sliding)

  • Between A and B: N₁ = W_B = 50 N, so F₁ = 0.35 × 50 = 17.5 N
  • Between table and A: N₂ = W_A + W_B = 150 N, so F₂ = 0.25 × 150 = 37.5 N
  • Total friction on A opposing motion = 55 N

Equations of motion (a = acceleration of A and C)

  • Block C: 500 − T = (500/9.81) a
  • Block A: T − 17.5 − 37.5 = (100/9.81) a

Adding: 500 − 55 = (600/9.81) a, so a = 445 × 9.81/600 = 7.28 m/s²

Tension in the rope AC: T = 500 − (500/9.81)(7.276) = 129.2 N (Rope force on B equals the friction 17.5 N.)

Answer: acceleration of block A = 7.28 m/s² (to the right). Driving force 500 N exceeds the friction of 55 N, so motion does start.

✓ Answer checked by Claude Opus 5.5 (high effort)

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