Moments and Equilibrium of Bodies

Lessons

Senior 4 · Physics

Moments and Equilibrium of Bodies

4 lessons

Lesson 1 of 4

Forces as Vectors and the Moment of a Force

25 min

Why the handle is far from the hinge

Push a classroom door near its hinges and it hardly moves; push the same amount at the handle and it swings open. A mechanic loosening a tight wheel nut on a minibus slips a pipe over the spanner to make it longer. A small child can lift a heavier friend on a see-saw by sitting farther from the middle. In each case a force has a turning effect that depends not only on how big the force is but on where and in what direction it acts. In S1 you met moments in simple cases; this lesson treats forces as vectors and gives the turning effect a precise definition.

Forces as vectors

A force has a size (magnitude) and a direction, so it is a vector. Forces acting at one point add by the parallelogram (or triangle) rule, not by simple arithmetic. The easiest way to add several forces is to resolve each into perpendicular components. A force $F$ at angle $\theta$ to the $x$ axis has components

F_x = F\cos\theta, \qquad F_y = F\sin\theta

Add all the $x$ components and all the $y$ components separately; the resultant has

R = \sqrt{R_x^2 + R_y^2}, \qquad \tan\varphi = \frac{R_y}{R_x}

where $\varphi$ is the angle of $R$ to the $x$ axis.

Figure 1.1: Adding two forces: the parallelogram rule and components give the same resultant.

Definition

The resultant of a set of forces is the single force that has the same effect as all of them together. To resolve a force is to replace it by two perpendicular forces (its components) that together have the same effect. Example: a 10 N force at 30° to the horizontal has components $10\cos 30^\circ = 8.7\,\mathrm{N}$ horizontally and $10\sin 30^\circ = 5.0\,\mathrm{N}$ vertically.

Worked example 1: pulling a canoe ashore

Two people pull a canoe up a beach with ropes. One pulls with 300 N at 20° to one side of the canoe's axis (the $x$ axis), the other with 250 N at 30° to the other side (Figure 1.1). Find the resultant force.

R_x = 300\cos 20^\circ + 250\cos 30^\circ = 281.9 + 216.5 = 498.4\,\mathrm{N}

R_y = 300\sin 20^\circ - 250\sin 30^\circ = 102.6 - 125.0 = -22.4\,\mathrm{N}

R = \sqrt{498.4^2 + 22.4^2} = 499\,\mathrm{N}, \qquad \varphi = \tan^{-1}\frac{-22.4}{498.4} = -2.6^\circ

The resultant is 499 N, turned 2.6° from the axis towards the 250 N rope, because that rope's sideways component (125.0 N) is the larger. Notice that it is less than $300 + 250 = 550\,\mathrm{N}$ : the sideways components partly cancel.

The moment of a force

A force can make a body turn about a point or axis. Its turning effect, the moment (or torque), is the force multiplied by the perpendicular distance from the pivot to the force's line of action (Figure 1.2):

M = F\,d = F\,r\sin\theta

Here $r$ is the distance from the pivot to the point where the force acts and $\theta$ the angle between $r$ and $F$ . The unit is the newton metre (N m). The moment is greatest when the force is at right angles to $r$ ( $\sin 90^\circ = 1$ ), and zero when the line of action passes through the pivot ( $d = 0$ ): pulling a door straight away from its hinges cannot open it.

Figure 1.2: Only the perpendicular distance d from the pivot to the line of action counts.

Moments have a sense: clockwise or anticlockwise. More precisely, torque is a vector, $\vec{\tau} = \vec{r} \times \vec{F}$ , of size $rF\sin\theta$ , pointing along the axis of rotation in the direction given by the right-hand rule (curl the fingers of your right hand in the sense of rotation; your thumb gives the direction). For forces in one plane we simply call anticlockwise moments positive and clockwise ones negative.

Definition

The moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action of the force: $M = F d$ . Its SI unit is the newton metre (N m). Example: a 40 N force applied at right angles to a door 0.80 m from the hinge has a moment of $40 \times 0.80 = 32\,\mathrm{N\,m}$ about the hinge.

Worked example 2: a spanner at an angle

A mechanic pushes with 80 N on the end of a spanner 0.25 m long. Find the moment about the nut when the force is (a) at 90° to the spanner, (b) at 60° to the spanner, as in Figure 1.2.

\text{(a)}\;\; M = 80 \times 0.25 = 20\,\mathrm{N\,m} \qquad \text{(b)}\;\; M = 80 \times 0.25 \times \sin 60^\circ = 17\,\mathrm{N\,m}

To get the most turning effect for the least effort, push at right angles to the spanner, as far from the nut as possible. A pipe that doubles the length doubles the moment.

The principle of moments

A body that is not rotating, or is rotating at a steady rate, has no unbalanced turning effect on it. This gives the principle of moments:

\text{for a body in rotational equilibrium:}\quad \sum M_{\text{clockwise}} = \sum M_{\text{anticlockwise}} \quad \text{about any point}

Experiment 1.1: testing the principle of moments

Apparatus: a metre rule, a knife edge (or a pencil taped to the bench), slotted masses with thread loops.

Figure 1.3: A balanced metre rule. Each weight is hung by a thread loop so that its distance from the pivot can be read exactly.

Balance the rule on the knife edge on its own and note the balance point (close to 50 cm, the centre of gravity of the rule).
Hang a 200 g mass at the 20 cm mark. Move a 150 g mass along the other side until the rule balances again. Record both distances from the pivot.
Repeat with other masses and positions. Each time, compare $m_1 g\,d_1$ with $m_2 g\,d_2$ .
Extension: move the pivot to the 40 cm mark and balance the rule with a single known mass on the short side. Use the principle of moments to find the mass of the rule itself.

Worked example 3: balancing the rule

(a) In Figure 1.3 the rule is balanced about its centre. Check that the principle of moments holds.

\text{anticlockwise} = 0.200 \times 9.8 \times 0.300 = 0.588\,\mathrm{N\,m}, \qquad \text{clockwise} = 0.150 \times 9.8 \times 0.400 = 0.588\,\mathrm{N\,m}

(b) The pivot is moved to the 40.0 cm mark. A 50.0 g mass at the 15.0 cm mark now balances the rule. The rule's weight acts at its centre of gravity, the 50.0 cm mark. Find the mass of the rule.

0.0500\,g \times (40.0 - 15.0) = M g \times (50.0 - 40.0) \;\Rightarrow\; M = \frac{0.0500 \times 25.0}{10.0} = 0.125\,\mathrm{kg} = 125\,\mathrm{g}

The distances could stay in centimetres here because they appear on both sides and cancel; $g$ cancels too.

Worked example 4: a see-saw

Uwase (30 kg) sits 1.6 m from the pivot of a see-saw. Where must her brother Mugisha (40 kg) sit to balance her? Take $g = 9.8\,\mathrm{N\,kg^{-1}}$ .

30 \times 9.8 \times 1.6 = 40 \times 9.8 \times x \;\Rightarrow\; x = \frac{30 \times 1.6}{40} = 1.2\,\mathrm{m}

The heavier child sits nearer the pivot. The see-saw itself, pivoted at its centre, adds no moment because its weight acts through the pivot.

Warning

Common mistakes. (1) Multiplying the force by the distance along the body instead of the perpendicular distance to the line of action. (2) Adding forces at an angle as plain numbers: resolve them first. (3) Writing the moment's unit as joules: N m is used for moments; the joule is reserved for energy. (4) Forgetting the weight of the beam or rule itself, which acts at its centre of gravity.

Tip

Take moments about a point where an unknown force acts: that force then has zero moment and drops out of the equation, leaving fewer unknowns.

The principle of moments (Physics Online). Watch how the moment is calculated from the perpendicular distance and how clockwise and anticlockwise moments are balanced in A-level problems.

✦

Practice

A force of 50 N acts at 35° above the horizontal. Find its horizontal and vertical components.
Forces of 6.0 N east and 8.0 N north act on a point. Find the size and direction of the resultant.
A door handle is 0.75 m from the hinges. A force of 30 N is applied at the handle (a) at right angles to the door, (b) at 40° to the door. Find the moment in each case.
A 2.0 m uniform plank pivoted at its centre has a 60 N weight 0.50 m from one end. What force applied at the other end, at right angles to the plank, balances it?
A metre rule pivoted at the 30.0 cm mark balances when a 100 g mass hangs at the 5.0 cm mark. Find the mass of the rule.
Three forces act on a ring: 20 N along $+ x$ , 15 N along $+ y$ , and a third force $P$ . The ring is in equilibrium. Find the size and direction of $P$ .
A student claims: "A force always produces a larger moment if it is applied farther from the pivot." Evaluate this claim.

Show answersHide answers

$50\cos 35^\circ = 41\,\mathrm{N}$ horizontally; $50\sin 35^\circ = 29\,\mathrm{N}$ vertically.
$R = \sqrt{6.0^2 + 8.0^2} = 10\,\mathrm{N}$ , at $\tan^{-1}\frac{8.0}{6.0} = 53^\circ$ north of east.
(a) $30 \times 0.75 = 22.5\,\mathrm{N\,m} \approx 23\,\mathrm{N\,m}$ ; (b) $30 \times 0.75 \times \sin 40^\circ = 14\,\mathrm{N\,m}$ .
The weight is 0.50 m from the pivot; the end is 1.0 m from it: $60 \times 0.50 = F \times 1.0$ , so $F = 30\,\mathrm{N}$ .
$100 \times (30.0 - 5.0) = M \times (50.0 - 30.0)$ , so $M = \frac{2500}{20.0} = 125\,\mathrm{g}$ .
$P$ must cancel the resultant of the other two: $|P| = \sqrt{20^2 + 15^2} = 25\,\mathrm{N}$ , directed opposite to it, at $\tan^{-1}\frac{15}{20} = 37^\circ$ below the $- x$ axis (that is, pointing into the third quadrant).
Not always. The moment is $Fr\sin\theta$ : what matters is the perpendicular distance to the line of action. A force applied far from the pivot but directed almost through it (small $\theta$ ) can have a smaller moment than one applied nearer the pivot at right angles. The claim holds only if the angle is the same.

Quick check

Key points

Forces are vectors: resolve into components $F\cos\theta$ and $F\sin\theta$ , add the components, then $R = \sqrt{R_x^2 + R_y^2}$ .
Moment $M = Fd = Fr\sin\theta$ (N m), where $d$ is the perpendicular distance from the pivot to the line of action; torque $\vec{\tau} = \vec{r} \times \vec{F}$ .
Principle of moments: in rotational equilibrium, total clockwise moment = total anticlockwise moment about any point.

Sources and credits

9.2: The Second Condition for Equilibrium, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: Playground seesaw.jpg by Killarnee, Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: The Principle of Moments - A Level Physics, Physics Online, embedded from YouTube.

The three diagrams, the experiment, the worked examples and the practice exercises are written by Ganzaa.

1Why the handle is far from the hinge

2Forces as vectors

3The moment of a force

4The principle of moments

Experiment 1.1: testing the principle of moments

Why the handle is far from the hinge

Forces as vectors

The moment of a force

The principle of moments