An object is in equilibrium when the resultant force on it is zero. To use this, balance the forces separately along two perpendicular directions, usually up and down, then left and right.
This lesson is part of SPM Physics force and motion II. It depends on resolving forces into components and combining forces and finding a resultant.
What is the balance-the-components routine?
Follow these steps.
- Draw the object and every force acting on it.
- Resolve any angled force into two perpendicular components.
- Write “up = down” and “left = right”, using the components.
- Solve for the unknowns.
The routine gives two equations, so it can find two unknown forces, such as two tensions.
Worked example 1: a hanging lamp
A lamp of weight 20 N hangs from two identical strings. Each string makes 30° with the horizontal. Find the tension in each string.
The tensions are equal by symmetry, so the horizontal components cancel. The vertical component of each tension is T sin 30° = 0.5T. Balance the vertical forces: 2 × 0.5T = 20, so T = 20 N.
Each string carries a tension of 20 N, which equals the whole weight, and not half of it. The strings are fairly flat, so a large part of each tension pulls sideways.
Worked example 2: a block at rest on a slope
A 6.0 kg block rests on a slope of 30°. Take g = 10 m s⁻², so the weight is 60 N.
Along the slope: the weight component down the slope is 60 sin 30° = 30 N, so the friction up the slope is 30 N. Perpendicular to the slope: the normal reaction equals 60 cos 30° = 52 N.
This question only needs the balance of forces. The friction does not have to be at its maximum for the block to stay at rest.
The mistake to avoid
The common mistake is to set each tension equal to half the weight, whatever the angles. The table shows how the tension changes with the angle to the horizontal.
| Angle to horizontal | Tension in each string (weight 20 N) |
|---|---|
| 90° (vertical) | 10 N |
| 30° | 20 N |
| 10° | 57.6 N |
The flatter the strings, the larger the tension. The half-weight rule works only when the strings are vertical. The units and significant figure checker helps you keep answers in newtons.
Check yourself
A picture frame of weight 12 N hangs from two strings. Each string makes 60° with the vertical. Find the tension in each string.
Answer
Each string makes 60° with the vertical, so its vertical component is T cos 60° = 0.5T. Balance the vertical forces: 2 × 0.5T = 12, so T = 12 N.
The horizontal components, each T sin 60°, are equal and opposite, so they cancel.
What to study next
Go on to using Hooke’s law and elastic energy, where a spring supplies the balancing force. The force and motion II practice set includes hanging and slope questions.
If you want a teacher to watch your component balance with you, see online one-to-one Physics tuition.