A uniform horizontal beam of length 2 m and weight 100 N rests on vertical supports at its two ends. A pair of equal opposite forces, each 20 N and separated by perpendicular distance 0.50 m, applies a clockwise couple to the beam. (a) Calculate the couple's torque. (b) Calculate both upward support reactions. (c) Explain why the couple changes the reactions but not their sum.

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Torque 10 N m; left reaction 45 N; right reaction 55 N; their sum stays 100 N. The torque of a couple is one force multiplied by the perpendicular separation between the two force lines: C = Fd = 20 × 0.50 = 10 N m. Do not double this using the full separation twice. A pure couple has zero resultant force, so vertical equilibrium gives Rleft + Rright = 100 N. The beam's weight acts at its midpoint because the beam is uniform. Take moments about the left support. The right reaction provides anticlockwise torque, so Rright × 2 = 100 × 1 + 10. This gives Rright = 55 N. Then Rleft = 100 − 55 = 45 N. The added clockwise turning effect redistributes support forces while its zero resultant leaves their total unchanged. Both answers are positive, consistent with the supports remaining in contact with the beam.

Supporting a beam with an applied couple 1

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