Solving Beam Problems with Moments | CAIE 9702 Paper 1 Notes

Paper 1

Physics9702

Solving Beam Problems with Moments

A beam question is one careful application of the principle of moments: mark every force, choose a pivot and balance the two turning directions.

The beam method

Draw the beam. Mark every force and its distance from the pivot.
Put the beam's own weight at its centre of gravity: the midpoint if it is uniform, the marked point if it is not.
Sort the moments into clockwise and anticlockwise.
Set the two sums equal and solve.

Worked example

Exam version: three people on a beam

A uniform beam is pivoted at its midpoint. A 60 kg person sits 3.0 m left of the pivot and an 80 kg person sits 3.0 m right of it. A third person of mass 45 kg sits at distance x, left of the pivot, to balance the beam. Find x.

The beam is uniform and pivoted at its midpoint, so its own weight acts at the pivot and has no moment.
Balance: $80g \times 3.0 = 60g \times 3.0 + 45g \times x$ .
Every term has $g$ , so it cancels: $240 = 180 + 45x$ , giving $x = \dfrac{60}{45} = 1.3 \text{ m}$ .

Answer

x = 1.3 \text{ m}

Working with weights on both sides lets g cancel. Working with masses from the start gives the same answer here, but write moments as force × distance in your working.

Common mistake

On a non-uniform beam, the weight does not act at the middle. Put it at the centre of gravity the question gives you. Reversed questions use this: they tell you the balance point and ask where the centre of gravity is.

Drag the load on this beam and watch both moment totals until they match:

Moment balance

The left load is fixed: 30 N, 2.0 m out, so it turns the beam anticlockwise with a moment of 60 N m. Move the right load until the beam sits level.

Right load20 NDistance2.0 m

Anticlockwise

30 × 2.0 = 60 N m

Clockwise

20 × 2.0 = 40 N m

Beam

tips anticlockwise

A side view of a bent arm. The forearm is horizontal, the biceps is visible and the hand holds a small ball. — The elbow is the pivot. The biceps acts very close to it, while the forearm weight and the ball act much farther away. The exact force-and-distance model is shown next.

Exact lever model. Take moments about the elbow: the hinge force has zero moment, so the small biceps distance must be balanced by the two larger load distances.

The same method explains your own arm. The biceps pulls up only 4.0 cm from the elbow, while a 50 N object in the hand sits 35 cm away, and the 15 N forearm weight acts at 16 cm.
Moments about the elbow: $F \times 0.040 = (15)(0.16) + (50)(0.35)$ , so $F = \dfrac{19.9}{0.040} \approx 500 \text{ N}$ . The muscle force is far greater than the ball's weight because the muscle acts through a much shorter distance from the pivot.

Your turnquick check

A 40 N load acts 0.30 m to the right of a pivot. Where should a 20 N load act on the left to balance it?

Show worked answer

Equate opposite moments: $40(0.30)=20x$ .
$x=0.60\ \text{m}$ to the left of the pivot.