Week 6 · Parallel coplanar forces: an unknown force from the resultant, sums of moments of parallel forces, and the resultant of several parallel forces · 3 lessons
1
Finding one of two parallel forces from the resultant, the other force and the distance, when the known force and the resultant point the same way and when they point opposite ways
Applied Mathematics Sec 3 Term 1, Unit 2, Lesson 1, Part 2
GoalFind an unknown parallel force and the distance between the lines of action from the resultant and one force, whether the forces point the same way or opposite ways.
New words
resultant · the one force that has the same effect as all the forces togetherline of action · the straight line along which a force pushes or pullslike and unlike forces · forces pointing the same way, and forces pointing opposite ways
Explanation
Two parallel forces F₁ and F₂ have a resultant R that is parallel to them too. Three facts tie the three forces together. First, choose the direction of R as the positive direction; then R = F₁ + F₂, and a force that points the other way is written with a minus sign. Second, let R act along a line through the point C. R has no turning effect about a point of its own line, so the moments of F₁ and F₂ about C must cancel: F₁ × d₁ = F₂ × d₂, where d₁ and d₂ are the distances from C to the two lines of action, and the two forces turn the rod in opposite senses. Third, the larger force is the nearer one. So the unknown force is F₂ = R − F₁ and the unknown distance is d₂ = F₁d₁ / F₂. If F₁ points the same way as R, then F₂ is smaller than R, points the same way too, and lies on the other side of R from F₁. If F₁ points against R, then F₂ = R + F₁ is bigger than R: the forces are unlike, they lie on the same side of R, and the larger force F₂ is the nearer one, so R acts outside the two forces on the side of the larger.
Worked examples
Two parallel forces act on a light rod. One is 30 N, the resultant is 50 N, and the line of action of the 30 N force is 12 cm from the resultant. Take the direction of R as positive (upwards). (a) The 30 N force points the same way as R. Find the other force F and the distance d of its line from R. Solution: F = 50 − 30 = 20 N, upwards. Moments about C, the point where R acts: 30 × 12 = 20 × d, so d = 18 cm, on the other side of R. The forces are 12 + 18 = 30 cm apart.
The same data, but (b) the 30 N force points against R, so up is positive and the 30 N force counts as −30. Solution: F = 50 − (−30) = 80 N, upwards. Moments about C: 30 × 12 = 80 × d, so d = 4.5 cm. Both forces lie on the same side of R, and the larger force (80 N) is the nearer. The forces are 12 − 4.5 = 7.5 cm apart. The picture shows the 30 N force pointing down.
Class activity
Balance a 30 cm ruler on a pencil. Put one coin on the 5 cm mark and a stack of two identical coins on the 20 cm mark. Slide the pencil until the ruler balances: it balances at the 15 cm mark. The pencil is where the resultant of the two weights acts. Check with the rule: 1 × 10 = 2 × 5 (distance of each coin from the pencil). What changes if you use three coins on the 20 cm mark?
Practice
1
Two parallel forces point the same way; one is 24 N and the resultant is 40 N, whose line is 10 cm from the 24 N force. Take the direction of R as positive. Find the other force and the distance between the lines of action.
F = 40 − 24 = 16 N. Moments about the point of R: 24 × 10 = 16 × d, so d = 15 cm on the other side: the forces are 25 cm apart.
2
Same resultant (40 N) and the same 10 cm, but now the 24 N force points against R. Take the direction of R as positive. Find the other force and the distance between the lines.
F = 40 − (−24) = 64 N in the direction of R. 24 × 10 = 64 × d gives d = 3.75 cm, on the same side as the 24 N force, so the forces are 10 − 3.75 = 6.25 cm apart.
3
A light rod AB carries an 18 N upward force at A. The resultant of this force and a force at B is 30 N upward and acts 20 cm from A. Up is positive. Find the force at B and the length AB.
F = 30 − 18 = 12 N upwards. Moments about the point of R: 18 × 20 = 12 × d, so d = 30 cm and AB = 20 + 30 = 50 cm.
4
An upward force of 90 N and an unknown parallel force have a resultant of 60 N downwards. The line of the resultant is 10 cm from the 90 N force. Take downwards (the direction of R) as positive. Find the unknown force and the distance between the lines.
The 90 N force counts as −90, so F = 60 − (−90) = 150 N downwards. 90 × 10 = 150 × d gives d = 6 cm, on the same side as the 90 N force, so the lines are 10 − 6 = 4 cm apart.
5
Can two unlike forces of 15 N and 10 N have a resultant of 30 N? Explain.
No. For unlike forces the resultant is the difference, 15 − 10 = 5 N, so it can never be larger than the larger force. Resultant 30 N would need a second force of 30 − 15 = 15 N (like forces) or 30 + 15 = 45 N (unlike), not 10 N.
