Equilibrium of concurrent forces and free body diagrams
Concurrent forces are in equilibrium when their vector sum is zero. A free body diagram shows every force on one body; then apply ΣF = ma.
Equilibrium of concurrent forces
Forces that act at the same point at the same time are called concurrent forces. They are in equilibrium when their resultant is zero. The body then has no acceleration: it stays at rest or keeps moving uniformly. Take three forces F₁, F₂ and F₃ at a point P. Find the resultant of F₁ and F₂ by the parallelogram law. For equilibrium, this resultant must be equal and opposite to F₃. So F₁ + F₂ + F₃ = 0. Examples: a lamp hanging still from two strings, and a tug of war in which neither team moves.
Worked example: the third force for equilibrium
Two forces act at a point: 6 N towards the east and 8 N towards the north. Find the third force that keeps the point in equilibrium.
- The two forces are at right angles. Their resultant R = √(6² + 8²) = √(36 + 64) = √100 = 10 N.
- Its direction: tan θ = 8/6 = 1.33, so θ ≈ 53° north of east.
- For equilibrium, the third force must be equal and opposite to R.
A force of 10 N pointing about 53° south of west (opposite to the resultant)
How to draw a free body diagram and solve a force problem
- 1Draw a simple, neat sketch of the whole situation.
- 2Choose the body you want to study. Imagine it cut free from everything around it, and draw it alone as a box or a dot. This is the free body.
- 3★Draw every external force on it as an arrow touching the body. The usual ones are weight mg (down), the normal force Fₙ (at right angles to the surface), tension T (along the string, away from the body), friction f (along the surface, against the motion) and any applied force F.
- 4Choose axes: one along the motion (or the likely motion) and one at right angles to it.
- 5Apply the second law along each axis: ΣFx = m ax and ΣFy = m ay.
- 6You need as many equations as unknowns. Solve them, then check the answer in an extreme case, such as m₁ = m₂.
★ key step
Worked example: two blocks joined by a string
Blocks of 3 kg and 2 kg lie on a smooth level floor, joined by a light horizontal string. A 15 N force pulls the 2 kg block forward, and the 3 kg block follows behind. Find the acceleration and the tension in the string.
- Vertically, for each block, the normal force balances the weight.
- Both blocks together: only the 15 N force acts horizontally. 15 N = (3 + 2) kg × a, so a = 3 m s⁻².
- FBD of the 3 kg block: only the tension T acts horizontally. T = m₁a = 3 kg × 3 m s⁻² = 9 N.
- Check with the 2 kg block: F − T = m₂a gives 15 − 9 = 6 N = 2 kg × 3 m s⁻². It agrees.
Acceleration 3 m s⁻²; tension 9 N (less than the applied 15 N)
Worked example: two masses over a pulley
Masses of 5 kg and 3 kg hang from the ends of a light string over a light, frictionless pulley. Find their acceleration and the tension in the string. Take g = 9.8 m s⁻².
- The 5 kg mass moves down and the 3 kg mass moves up, both with acceleration a. The tension T is the same on both sides.
- FBD of the 5 kg mass: m₁g − T = m₁a.
- FBD of the 3 kg mass: T − m₂g = m₂a.
- Add the two: (m₁ − m₂)g = (m₁ + m₂)a, so a = (5 − 3) × 9.8 ÷ 8 = 2.45 m s⁻².
- T = m₂(g + a) = 3 × (9.8 + 2.45) = 3 × 12.25 = 36.75 N.
- Check with T = 2m₁m₂g/(m₁ + m₂) = 2 × 5 × 3 × 9.8 ÷ 8 = 36.75 N.
a = 2.45 m s⁻²; T ≈ 36.8 N
How to draw the free body diagrams for the pulley problem
- Draw a small circle for the pulley, hung from the ceiling by a short line.
- Draw a string over it, with the heavier block m₁ on the left and the lighter block m₂ on the right.
- Draw m₁ again on its own as a box: an arrow T pointing up and an arrow m₁g pointing down. Beside it, write a ↓.
- Draw m₂ on its own: an arrow T up and an arrow m₂g down. Beside it, write a ↑.
- Label the tension with the same letter T on both sides, because the string is light and the pulley frictionless.
Engineers draw free body diagrams before they design a lift, a crane or a bridge, so that every force is balanced. A ceiling fan hanging still is a simple case of forces in equilibrium.
5 marks: “Two masses m₁ and m₂ (m₁ > m₂) hang from the ends of a light string over a frictionless pulley. Draw their free body diagrams and find the acceleration and the tension. Find the values for m₁ = 5 kg and m₂ = 3 kg. (g = 9.8 m s⁻²)” FBDs: T up and m₁g down on m₁ (a down); T up and m₂g down on m₂ (a up). m₁g − T = m₁a and T − m₂g = m₂a. Adding: a = (m₁ − m₂)g/(m₁ + m₂); T = 2m₁m₂g/(m₁ + m₂). Values: a = 2 × 9.8 ÷ 8 = 2.45 m s⁻²; T = 30 × 9.8 ÷ 8 ≈ 36.8 N.
2 marks: “What are concurrent forces? State the condition for three concurrent forces to be in equilibrium.” Forces acting at the same point at the same time. For equilibrium their vector sum must be zero: F₁ + F₂ + F₃ = 0, so the resultant of any two is equal and opposite to the third.
3 marks: “An 8 kg load is pulled up by a rope with an acceleration of 1.5 m s⁻². Find the tension in the rope. (g = 9.8 m s⁻²)” FBD: T up, mg down; net force up = ma. T − mg = ma, so T = m(g + a) = 8 × (9.8 + 1.5) = 8 × 11.3 = 90.4 N.
Cut the body free, draw every arrow, then ΣF = ma along each axis.