⚖️ Rule 1: the forces balance
The beam does not move up or down, so the upward forces equal the downward force.
force at A + force at B = weight of the object
This is true wherever the object sits.
Forces, Movement, Shape and Momentum » Forces on a Supported Beam
What you'll learn this session
Study time: 30 minutes
Edexcel spec: 1.33P
- Why the two upward forces on a supported beam always add up to the weight of the object on it
- How the two forces change as a heavy object moves along the beam
- How to calculate each support force by taking moments about the other support
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Unlock This CourseA beam held up at both ends
This lesson is examined in Paper 2 only.
Think of a plank laid across a stream as a simple bridge. Each end rests on a bank. When you stand on the plank, your weight pushes down and the two banks push up on the ends of the plank. This lesson is about those two upward forces and how they share the load.
To picture it, draw a long thin horizontal rectangle for the beam. Put a small triangle under each end for the supports and label them A (left) and B (right). Draw an arrow pointing straight up at each support. Draw one arrow pointing straight down from the heavy object, and label it with the object's weight.
Exam questions call the beam a light beam. This tells you the weight of the beam is so small compared with the heavy object that you can ignore it. Only three forces matter: the weight of the object and the two upward forces from the supports.
Key terms:
- Light beam: a beam whose own weight is small enough to ignore.
- Support force: the upward force that a support exerts on the beam.
Two rules for a beam that stays still
The beam is not moving up or down, and it is not turning. That gives two rules.
🔄 Rule 2: the moments balance
The beam does not turn, so you can use the principle of moments. The clever part is that you may choose either support as the pivot.
Rule 1 tells you the total. Rule 2 tells you how that total is shared between the two supports.
How the forces change with position
The position of the heavy object decides how the load is shared.
- Object in the middle: the two upward forces are equal. Each one is half the weight of the object.
- Object moved towards one support: the force at that support increases and the force at the other support decreases. The total stays the same.
- Object directly above one support: that support provides all of the upward force. The force at the other support is zero.
A walker of weight 600 N crosses a light footbridge 6.0 m long, from support A to support B. The table shows the two upward forces at five positions.
| Distance of walker from A (m) | Force at A (N) | Force at B (N) | Total upward force (N) |
|---|---|---|---|
| 0 | 600 | 0 | 600 |
| 1.5 | 450 | 150 | 600 |
| 3.0 | 300 | 300 | 600 |
| 4.5 | 150 | 450 | 600 |
| 6.0 | 0 | 600 | 600 |
Look at the pattern. The force at B rises steadily as the walker gets closer to B. The force at A falls by the same amount each time. The last column never changes.
Calculating the support forces
There are two unknown forces, which looks like a problem. The trick is to take moments about one of the supports. The force at that support acts at the pivot, so its moment is zero and it drops out of the equation. That leaves only one unknown.
- Choose one support as the pivot, for example A.
- Write: sum of clockwise moments = sum of anticlockwise moments.
- The weight of the object has a moment about A. Its distance is measured from A to the object.
- The force at B has a moment the other way. Its distance is the full length of the beam.
- Solve for the force at B. Then use Rule 1 to find the force at A.
Worked example 1: finding both forces
A light plank 4.0 m long rests on two supports, A at the left-hand end and B at the right-hand end. A painter of weight 800 N stands on the plank, 1.0 m from A. Calculate the upward force at each support.
Take moments about A. The painter's weight turns the plank clockwise about A. The force at B turns it anticlockwise.
sum of clockwise moments = sum of anticlockwise moments
800 × 1.0 = force at B × 4.0
800 = force at B × 4.0
force at B = 800 / 4.0
force at B = 200 N
force at A + force at B = weight of the painter
force at A = 800 − 200
force at A = 600 N
Does this make sense? The painter is closer to A, so A should provide the larger force. It does. You can also check by taking moments about B. The painter is 3.0 m from B, so force at A × 4.0 = 800 × 3.0, which gives 2400 / 4.0 = 600 N again.
Notice the shares. The painter is a quarter of the way along from A, and the far support B provides a quarter of the weight. The near support provides the other three quarters.
Worked example 2: finding the position
A light beam 2.0 m long rests on a support at each end. A crate of weight 300 N is placed on the beam. The upward force at the left-hand support, A, is 210 N. Calculate the distance of the crate from A.
force at B = 300 − 210 = 90 N
Take moments about A. Call the distance of the crate from A, d.
sum of clockwise moments = sum of anticlockwise moments
300 × d = 90 × 2.0
300 × d = 180
d = 180 / 300
d = 0.60 m
The crate is closer to A than to B, which fits with A providing the larger force.
Why bridges care
A heavy lorry driving across a bridge is a heavy object moving along a supported beam. As the lorry drives on, the support at the near end carries almost all of its weight. Half way across, the two supports share it equally. As it leaves, the far support carries almost all of it. So each support must be strong enough to hold nearly the whole weight of the heaviest vehicle, not just half of it. A real bridge is not a light beam, so each support also has to hold up its share of the weight of the bridge itself.
Common mistakes
- Giving the larger force to the wrong support. The support nearer the object always provides the larger force. Check this at the end of every calculation.
- Using the wrong distance. When you take moments about A, the object's distance is measured from A, and the distance for the force at B is the whole length of the beam.
- Saying the forces are always equal. They are only equal when the object is in the middle.
- Thinking the total changes. Moving the object changes how the load is shared. The two upward forces still add up to the weight of the object.
- Adding in a weight for the beam. If the question says the beam is light, ignore its weight.
Exam-style question
A light beam 3.0 m long rests on two supports. Support P is at the left-hand end and support Q is at the right-hand end. A motor of weight 900 N is placed on the beam, 1.0 m from P.
(a) State the total upward force on the beam from the two supports. [1 mark]
(b) Calculate the upward force on the beam from support Q. [3 marks]
(c) Calculate the upward force on the beam from support P. [1 mark]
(d) The motor is slowly moved along the beam towards Q. Describe what happens to the upward force at each support. [2 marks]
(e) Explain why it matters that the beam is described as light. [1 mark]
Model answer
(a) 900 N (1)
(b) Taking moments about P: clockwise moment = 900 × 1.0 = 900 Nm (1)
force at Q × 3.0 = 900 (1)
force at Q = 900 / 3.0 = 300 N (1)
(c) force at P = 900 − 300 = 600 N (1)
(d) The force at Q increases (1). The force at P decreases (1). (The two forces still add up to 900 N.)
(e) The weight of the beam is small enough to ignore, so the only downward force to include is the weight of the motor (1).
Exam tip
In part (b), take moments about P, the support you are not asked about. The force at P then has no moment and disappears from the equation. Before you move on, check your answers: the motor is nearer to P, so P must have the larger force, and 600 N + 300 N must equal 900 N.