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2-1 (f) The resultant of a system of parallel forces

2-1 (f) The resultant of a system of parallel forces

Point 6 of 12 — Parallel Coplanar Forces: Resultant and EquilibriumQuestion bank: 0 questions (0 with pictures)

06 2-1 (f) The resultant of a system of parallel forces

Procedure

  • Choose a positive direction (a unit vector c^ along the forces) and add the forces with signs: R=∑F.
  • If R≠0, take moments about a convenient point O (often the first force): R⋅x=∑F⋅d where x is the distance from O to the line of action of R, and d the signed distance of each force.
  • State: the magnitude, direction and position of the resultant (where it cuts the line through the points).
✍️ Worked examples

Example 6: A,B,C,D,E lie on a line with AB:BC:CD:DE=2:3:4:7. Like forces of 30, 50, 20, 70, 40 N act at A,B,C,D,E. Find the resultant.

Solution:

  • R=30+50+20+70+40=210 N, in the common direction.
  • Let AB=2x, so the distances from A are 0,2x,5x,9x,16x.
  • Moments about A: R⋅AO=50(2x)+20(5x)+70(9x)+40(16x)=100x+100x+630x+640x=1470x.
  • AO=1470x210=7x, so AOAE=7x16x=716.
  • The resultant divides AE internally in the ratio 7:16 from A.

Balance (centre of parallel weights)

  • The weights of particles on a light rod have a resultant equal to the total weight, acting at the balance point.
  • Example: a light rod 1 m long carries 2 N at one end and 3 N at the other; the balance point is 25×100=40 cm from the 3 N end.

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