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Force for just motion of block B

Short answer

From block A equilibrium, tension T = 4.851 N, normal N1 = 18.74 N, friction F1 = 4.69 N. For block B, ground normal N2 = 68.74 N, ground friction F2 = 17.19 N. The force P = F1 + F2 = 21.87 N, approximately 21.9 N.

Determine the force required for just motion of the block B shown in the figure below. Take μs = 0.25 for all surfaces. [Figure: block A (20 N) on block B (50 N); A tied to a wall by a string at 15°; B pulled by force P to the left on the ground.]

Reading of the figure: block A (20 N) sits on block B (50 N), which is on the ground. A is tied to a wall by a string at 15° to the horizontal (rising to the right). P pulls B to the left. μₛ = 0.25 for all surfaces. B slides to the left under A, which is held by the string.

Block A (held by the string)

Forces: weight 20 N, normal N₁ from B, friction F₁ = μN₁ from B (to the left, the direction of B's motion), tension T at 15°.

ΣFy = 0:  N1 + T sin15° = 20      ->  N1 = 20 − 0.2588 T
ΣFx = 0:  T cos15° = F1 = 0.25 N1
          0.9659 T = 0.25 (20 − 0.2588 T)
          (0.9659 + 0.0647) T = 5
          T = 4.851 N

N₁ = 20 − 0.2588 × 4.851 = 18.74 N F₁ = 0.25 × 18.74 = 4.69 N

Block B (just about to move to the left)

Forces: P to the left, weight 50 N, N₁ = 18.74 N pressing down from A, ground normal N₂, friction from A (F₁ = 4.69 N to the right) and ground friction F₂ (to the right).

N2 = 50 + 18.74 = 68.74 N
F2 = 0.25 × 68.74 = 17.19 N
P = F1 + F2 = 4.69 + 17.19 = 21.87 N

Answer: P ≈ 21.9 N.

✓ Answer checked by Claude Opus 5.5 (high effort)

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