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Force P to move block A (blocks A and B)

Short answer

For block B, string tension T = 5.905 N, normal from A N1 = 17.05 N, friction F1 = 0.3 N1 = 5.11 N. On block A, ground normal N2 = 40 + 17.05 = 57.05 N, ground friction F2 = 17.11 N. P = F1 + F2 = 22.23 N, so force required is about 22.2 N to the left.

Two blocks A and B of 40 N and 20 N respectively are in equilibrium position as shown in the figure below. Calculate the force P required to move block A. Take μ = 0.3 for all surfaces. [Figure: block B resting on block A, B tied to a wall by a string inclined at 30°; block A on the ground pulled by force P to the left.]

Reading of the figure: block B (20 N) rests on block A (40 N), which is on the ground. B is tied to a wall by a string inclined at 30° to the horizontal (rising towards the wall, on the right). P pulls A to the left. μ = 0.3 for all surfaces. When A moves left, B tends to move with it, but the string holds B, so B stays at rest and A slides under it.

Block B (equilibrium, friction from A at the limit)

Forces: weight 20 N, normal N₁ from A, friction F₁ = μN₁ from A (acting to the left, the direction of A's motion), string tension T (up and to the right at 30°).

ΣFy = 0:  N1 + T sin30° − 20 = 0     ->  N1 = 20 − 0.5 T
ΣFx = 0:  T cos30° − F1 = 0, F1 = 0.3 N1
          0.866 T = 0.3 (20 − 0.5 T)
          0.866 T + 0.15 T = 6
          T = 5.905 N

N₁ = 20 − 0.5 × 5.905 = 17.05 N F₁ = 0.3 × 17.05 = 5.11 N

Block A (moving to the left, so all frictions act to the right)

Forces: P to the left, weight 40 N, normal pressure N₁ = 17.05 N from B (downward), ground normal N₂, friction from B (F₁ = 5.11 N), ground friction F₂ = 0.3 N₂.

ΣFy = 0:  N2 = 40 + 17.05 = 57.05 N
F2 = 0.3 × 57.05 = 17.11 N
ΣFx = 0:  P = F1 + F2 = 5.11 + 17.11 = 22.23 N

Answer: P ≈ 22.2 N (to the left).

✓ Answer checked by Claude Opus 5.5 (high effort)

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