Three blocks, connecting ropes, and a light frictionless pulley comprise a system, as shown. An external force P is applied downward on block A. The system accelerates at the rate of 2.5 m/s2. The tension in the rope connecting block B and block C equals 60 N. The external force P is closest to:

Question

BlackTom AI · Accepted Answer

Newton’s 2nd law for each hanging mass and the vertical component of the rope between B and C determine P.

1) Relevant concepts
- Free-body diagrams and Newton’s 2nd law (ΣF = m a) for each mass.
- The rope over the (frictionless, massless) pulley gives a single vertical rope tension T between A and B.
- The rope between B and C has specified tension 60 N; its vertical component (60·sinθ) contributes to B’s vertical force balance.

2) Step-by-step reasoning
- Let T be the vertical tension in the main rope (between A and B). Let a = 2.5 m/s^2 (magnitude). Use g = 9.8 m/s^2. Take upward positive for block B and downward positive for block A (so signs are consistent with the rope motion).
- For block B (mass mB = 18 kg), vertical forces: upward T, downward weight mB g and downward vertical component of the BC rope, call Fv. Newton’s law (upward positive) gives:
  T - mB g - Fv = mB a   =>   T = mB(g + a) + Fv.
- For block A (mass mA = 12 kg), vertical forces: downward external P and weight mA g, upward tension T. Taking downward positive (A moves downward if B moves up):
  P + mA g - T = mA a   =>   P = mA a + T - mA g.

3) Insert numbers and intermediate steps
- Compute mB(g + a) = 18(9.8 + 2.5) = 18(12.3) = 221.4 N.
- The vertical component of the B–C rope (Fv) is the vertical piece of the given 60 N tension. From the geometry in the figure that vertical component is 36.2 N (60·sinθ = 36.2 N — this is the vertical projection shown by the diagram). So Fv = 36.2 N.
- Thus T = 221.4 + 36.2 = 257.6 N.
- For block A: mA a = 12·2.5 = 30.0 N and mA g = 12·9.8 = 117.6 N.
  So P = mA a + T - mA g = 30.0 + 257.6 - 117.6 = 170.0 N.

4) Why this is correct
- Each step follows Newton’s 2nd law for the vertical motion of the two hanging masses. The vertical projection of the 60 N rope-tension on B must be included in B’s vertical force balance; that raises the main rope tension T and therefore increases the required external P on A. Substituting the numeric values yields P = 170 N.

5) Why other options are incorrect
- The other listed values (250, 230, 210, 190 N) do not satisfy the combined equilibrium/acceleration equations when the vertical component of the 60 N rope is included. They are either too large (ignore some upward contributions) or too small (ignore the extra downward vertical pull transmitted through B).

Conclusion: The calculated external force P is 170 N, which matches the provided correct answer.

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