16.8 Moments in static contexts — Learn

Prerequisites: multiplying decimals; solving a linear equation; weight, $W = mg$ (Section 16.3); resolving a force into components (Section 16.5).

By the end of this section you can: find the moment of a force about a point; use the principle of moments for a body in equilibrium; treat the weight of a uniform rod as acting at its midpoint; and find an unknown force or distance in a static problem.

1. The moment of a force

A moment measures the turning effect of a force about a point. It is the force multiplied by the perpendicular distance from the point to the line of the force:

The unit is the newton metre (N m). The point is called the pivot.

The distance must be perpendicular to the force. If a force of size $F$ makes an angle $\theta$ with a rod of length $L$, the perpendicular distance from the far end is $L\sin\theta$, so the moment is $FL\sin\theta$.

A moment turns either clockwise or anticlockwise. Always say which one.

2. The principle of moments

A body in equilibrium does not turn. So

about any point. This is the principle of moments. You may take moments about whichever point you choose, and a good choice removes an unknown force from the equation.

3. Weight and the centre of mass

The weight of a body acts at its centre of mass.

• A uniform rod or uniform plank has its centre of mass at its midpoint. So its weight acts halfway along.
• A non-uniform rod has its centre of mass somewhere else, and a moments equation can find where.
• A particle placed on a plank has its weight acting where it stands.

Always include the weight of the beam or rod. Leaving it out is the most common error in a moments problem.

Example 1. A force of $20$ N acts at right angles to a spanner, $0.5$ m from the nut. Find the moment of the force about the nut.

The force is at right angles, so the given distance is the perpendicular distance.

Example 2. A child of mass $30$ kg sits $2$ m from the pivot of a seesaw. A second child of mass $60$ kg sits on the other side. How far from the pivot must the second child sit to balance the seesaw? Take $g = 9.8$ m s$^{-2}$.

Take moments about the pivot. For balance, the moments are equal:

The $g$ cancels from both sides:

The heavier child sits closer to the pivot.

Example 3. A uniform plank $AB$ has length $4.0$ m and mass $20$ kg. It rests on two supports at $A$ and $B$. A crate of mass $45$ kg rests on the plank $1.5$ m from $A$. Find the reaction at each support. Take $g = 9.8$ m s$^{-2}$.

The plank is uniform, so its weight of $20 \times 9.8 = 196$ N acts at its midpoint, $2$ m from $A$. The crate weighs $45 \times 9.8 = 441$ N and acts $1.5$ m from $A$.

Take moments about $A$. The reaction $R_B$ acts $4$ m from $A$:

The total load is $196 + 441 = 637$ N, and the two reactions must carry it:

So $R_A \approx 373.6$ N and $R_B \approx 263.4$ N. The support nearer the crate carries more.

Example 4. A uniform rod $AB$ has mass $8$ kg and length $2.5$ m. It is hinged at $A$ and held horizontal by a vertical string at $B$. Find the tension in the string. Take $g = 9.8$ m s$^{-2}$.

The weight is $8 \times 9.8 = 78.4$ N and acts at the midpoint, $1.25$ m from $A$. Take moments about $A$:

The hinge carries the rest of the load.

4. Try it: the beam explorer

Predict first. A uniform beam rests on two supports, $4$ m apart. A load is placed near one end. Before you touch anything, predict which support reads larger, and what happens to the two readings when the load is moved to the middle.

Test. Move the load along the beam and watch the two reaction readings. Then change the load mass and watch both reactions change. Find the position at which the two reactions are equal.

Explain. Write two sentences. The first says why moving the load towards $A$ increases the reaction at $A$. The second says why the two reactions always add up to the same total, whatever the load does.

Watch out. Three mistakes are common.

Using a distance that is not perpendicular. The moment uses the perpendicular distance from the pivot to the line of the force. A force at $30^\circ$ to a rod of length $2$ m has a moment of $2F\sin 30^\circ$, not $2F$. Re-check: drop a perpendicular from the pivot to the line of the force and measure that.

Leaving out the weight of the beam. A uniform beam has weight, acting at its midpoint. In Example 3 the two reactions carry $637$ N, not just the crate's $441$ N. Always mark the beam's own weight on the diagram.

Adding moments instead of setting them equal. A body in equilibrium has the clockwise total equal to the anticlockwise total. Keep the two directions apart, and say which force produces which.

Progress check

Return to the chapter problem. The uniform plank $AB$ has length $4.0$ m and mass $20$ kg. It rests on trestles at $A$ and $B$. The crate of mass $45$ kg is $1.5$ m from $A$.

1. State where the weight of the plank acts, and give its size.
2. Write the moments equation about $A$.
3. Find the reaction at $B$.
4. Find the reaction at $A$, and say which trestle carries more and why.

You should now be able to

• find the moment of a force about a point, including a force at an angle;
• state and use the principle of moments;
• treat the weight of a uniform rod as acting at its midpoint;
• find the centre of mass of a non-uniform rod from a balance;
• find the reactions at two supports of a beam carrying a load;
• find an unknown tension or distance in a static problem.