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Tension in cable holding a smooth rod

Short answer

Take moments about B to eliminate T and R_B: N_D × (3/sin70°) = W × 2cos70°. With W=98.1 N, N_D=21.02 N. Horizontal equilibrium gives T = N_D sin70° = 19.75 N.

Determine the tension in the cable BC which holds a part AB of length 4 m from sliding. The part has a mass of 10 kg. Assume all the contact surfaces are smooth. [Figure: rod AB of length 4 m leaning with B on the floor and inclined at 70° to the horizontal, resting at D on a 3 m high vertical wall edge; cable BC horizontal along the ground from B to C at the foot of the wall.]

Given (read from the figure): rod AB, L = 4 m, mass 10 kg, inclined at 70° to the horizontal. B is on the smooth floor, the rod rests on the smooth top edge D of a wall 3 m high, and the horizontal cable BC joins B to the foot C of the wall. Take g = 9.81 m/s².

Forces on the rod (FBD)

  • Weight W = 10 × 9.81 = 98.1 N at the mid-point of AB (2 m from B).
  • Reaction at B, R_B, vertical (floor smooth).
  • Reaction at D, N_D, perpendicular to the rod (smooth contact at an edge).
  • Tension T at B, horizontal along BC.

Geometry

BD = 3 / sin 70° = 3 / 0.9397 = 3.193 m

Moments about B (this removes T and R_B)

N_D × BD = W × 2 cos 70° N_D = 98.1 × 2 × 0.3420 / 3.193 = 21.02 N

Horizontal equilibrium

N_D acts perpendicular to the rod, so its horizontal component is N_D sin 70° and it pushes the foot B away from the wall. The cable must balance it: T = N_D sin 70° = 21.02 × 0.9397 = 19.75 N

(Check: vertically R_B = W − N_D cos 70° = 98.1 − 7.19 = 90.9 N, which is positive, so B stays on the floor.)

Answer: Tension in cable BC ≈ 19.75 N (about 19.8 N).

✓ Answer checked by Claude Opus 5.5 (high effort)

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