Concurrent forces and what equilibrium means
Concurrent forces are forces whose lines all meet at one point. A knot with three ropes is a good example: every rope pulls on the same knot.
A force is a push or a pull. We draw it as an arrow: the length shows how big it is, the arrow head shows the direction.
The object is in equilibrium when the forces on it cancel out and nothing is left over. Then the object is either at rest (static equilibrium) or moving at a steady speed in a straight line (a car on a straight road at constant speed, a lift going up smoothly). Both cases have no acceleration, because the net force is zero (Newton's first law).
The condition for equilibrium: ΣF = 0
Add all the force vectors. The sum is called the net force. For equilibrium:
ΣF = 0
Arrows point in different directions, so we split each slanted force into two parts, one along x (left-right) and one along y (up-down). This is called resolving the force. Then the rule becomes two simple rules:
- ΣFx = 0: forces to the right = forces to the left.
- ΣFy = 0: forces up = forces down.
For a force F at angle θ above the horizontal: Fx = F cos θ and Fy = F sin θ.
Two forces in equilibrium must be equal and opposite. Three forces in equilibrium, drawn tip to tail, form a closed triangle. Any one of them is equal and opposite to the sum of the other two.
Steps to solve an equilibrium problem
- Draw the object as a dot. Mark every force acting on it: weight, ropes, push, normal force, friction.
- Choose x and y axes. Pick them so most forces lie along an axis (on a slope, one axis along the slope).
- Resolve slanted forces into components.
- Write ΣFx = 0 and ΣFy = 0.
- Solve for the unknowns and check the answer makes sense.
Use SI units: force in newtons (N), weight W = mg, with g ≈ 9.8 m/s² (we use 10 m/s² for easy sums).
Application: a weight hanging from two ropes
A weight W hangs from a knot held by two equal ropes. Each rope makes angle θ with the horizontal. Sideways: the two horizontal pulls cancel. Up and down: 2T sin θ = W, so
T = W ÷ (2 sin θ)
When θ = 90° (ropes straight up) each rope carries W/2. As θ gets smaller, sin θ gets smaller and T gets bigger. When θ is very small the rope pulls enormously, and it can snap. A rope can never be perfectly straight with a weight at its middle, because sin 0° = 0 would need an infinite pull.
Application: a block on a slope, a sled, a pushed ball
Block on a smooth slope of angle α, held by a rope along the slope. Take one axis along the slope. The weight splits into W sin α along the slope (down) and W cos α into the slope. So rope pull T = W sin α and normal force N = W cos α.
Sled pulled at an angle at constant speed: the pull P at angle θ splits into P cos θ (forward) and P sin θ (up). Forward: P cos θ = friction. Up: N + P sin θ = W, so the ground pushes less than the weight.
Ball on a string pushed sideways by a horizontal force F so the string makes angle φ with the vertical: T cos φ = W and T sin φ = F, so F = W tan φ.
Key formulas and definitions
- Equilibrium: ΣF = 0 (vector sum of forces is zero)
- ΣFx = 0 and ΣFy = 0
- Components: Fx = F cos θ, Fy = F sin θ (θ from the x-axis)
- Weight: W = m g (g ≈ 9.8 m/s²)
- Two equal ropes at θ to the horizontal: T = W / (2 sin θ)
- Smooth slope, rope along slope: T = W sin α, N = W cos α
- Unit of force: newton (N)
Worked examples
1. A 5 kg box rests on a table. Find the normal force. (g = 10 m/s²)
Weight W = mg = 5 × 10 = 50 N down. At rest, ΣFy = 0, so normal force N = 50 N up.
2. A 12 N lamp hangs from two equal ropes, each at 60° to the horizontal. Find the tension in each rope.
Up and down: 2T sin 60° = 12. T = 12 ÷ (2 × 0.866) = 6.93 N. (The sideways pulls cancel.)
3. The same 12 N lamp, but now each rope is at 30° to the horizontal. Find T.
T = 12 ÷ (2 × sin 30°) = 12 ÷ (2 × 0.5) = 12 N. A flatter rope pulls harder, even though the lamp did not change.
4. A 20 N block rests on a smooth slope of 30°, held by a rope parallel to the slope. Find the rope tension and the normal force.
Along the slope: T = W sin 30° = 20 × 0.5 = 10 N. Into the slope: N = W cos 30° = 20 × 0.866 = 17.3 N.
5. A sled of weight 200 N is pulled at constant speed with a 50 N rope at 30° above the ground. Find the friction and the normal force.
Constant speed means equilibrium. Forward: friction = 50 cos 30° = 43.3 N. Up: N + 50 sin 30° = 200, so N = 200 − 25 = 175 N.
6. A 30 N weight hangs from a knot. One rope goes to the ceiling at 45° to the horizontal. Another rope pulls the knot horizontally to a wall. Find both tensions.
Up and down: T1 sin 45° = 30, so T1 = 30 ÷ 0.707 = 42.4 N. Sideways: T2 = T1 cos 45° = 42.4 × 0.707 = 30 N.
7. A 2 kg ball hangs on a string. A horizontal push F holds it so the string makes 30° with the vertical. Find F and the string tension. (g = 10 m/s²)
W = 20 N. T cos 30° = 20, so T = 20 ÷ 0.866 = 23.1 N. F = T sin 30° = 11.5 N (or F = W tan 30° = 20 × 0.577 = 11.5 N).
Common mistakes
- Thinking "net force zero" means the object must be at rest. It can also move at a steady speed in a straight line.
- Adding force magnitudes like plain numbers. Forces are vectors; direction matters, so use components.
- Forgetting that rope tension is bigger than the weight when the ropes are nearly flat.
- Using sin where cos is needed. Check which angle you were given, the one with the horizontal or the one with the vertical.