06 2-1 (f) The resultant of a system of parallel forces
Procedure
- Choose a positive direction (a unit vector along the forces) and add the forces with signs: .
- If , take moments about a convenient point (often the first force): where is the distance from to the line of action of , and 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: lie on a line with . Like forces of 30, 50, 20, 70, 40 N act at . Find the resultant.
Solution:
- N, in the common direction.
- Let , so the distances from are .
- Moments about : .
- , so .
- The resultant divides internally in the ratio from .
Example 7:
- Four forces act on points of a line with cm, cm, cm: 10 N down at , 25 N up at , 15 N down at , 20 N up at .
- Find the resultant.
Solution:
- Take upward as positive: N upward.
- Distances from : cm.
- Moments about : N·cm.
- , so cm from , on .
Example 8:
- A light rod carries a downward force 8 N at , an upward 14 N at 3 m from , and a downward 4 N at (6 m from ).
- Find the resultant.
Solution:
- Upward positive: N upward.
- Moments about : N·m.
- m from along produced beyond (outside the rod).
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 cm from the 3 N end.