Moment of a Force

The moment (or torque \( \tau \) ) of a force about a fixed point \(O\) is:

\[ \tau = Fd \]

where \(F\) is the force and \(d\) is the perpendicular distance from \(O\) to the line of action of the force.

Moment diagram

A turning moment twists the object and is also known as torque and has SI units Newton metres (N m),
which is the same unit used for work (1 Joule = 1 N m) but is not the same thing!

Vector Definition

Torque is the vector cross product:

\[ \vec{\tau} = \vec{r} \times \vec{F} \]

Where r is the position vector of the point at which the force is applied.

\[ \tau = |\mathbf{r}|\;|\mathbf{F}|\;\sin\theta \]

This can also be displayed as

\[ \tau = r_{\perp} F \]

Where \( r_{\perp} \)represents the perpendicular distance from the point to the force.

Magnitude:

\[ \tau = rF\sin\theta = Fr_\perp \]

Sign Convention

Anticlockwise moments are positive. Clockwise moments are negative.

\[ \tau^{+} = \overleftarrow{\text{anticlockwise}}\;\text{moment} \] \[ \tau^{-} = \overrightarrow{\text{clockwise}}\;\text{moment} \]

or showing final force directions as

ant2 and any

Examples

Calculate the moment of a 5 N force about point A:

Example diagram

\[ \tau = Fd \] \[ = 5 \times 2 \] \[ = 10\;\text{Nm anticlockwise} \] \[ \text{or} \] \[ \tau^{+} = 10\;\text{Nm} \]

Example 2

\[ \tau = Fd \] \[ = 5 \times 0.5 \] \[ = 2.5\;\text{Nm clockwise} \] \[ \text{or} \] \[ \tau^{-} = 2.5\;\text{Nm} \]

Example 4

Extending the perpendicular:

Perpendicular extension

\[ \tau = Fd \] \[ = 5 \times 20 \sin 30^\circ \] \[ = 50\;\text{Nm clockwise} \] \[ \text{or} \] \[ \tau^{-} = 50\;\text{Nm} \]

Pstraight line

\[ \tau = Fd \] \[ = 5 \times 20 \sin 0^\circ \] \[ = 0\;\text{Nm} \]

See-Saw Equilibrium

For a see-saw in equilibrium, the total moment about the pivot is zero.

\[ \sum \tau = 0 \]

\[ \tau^{+} = \tau^{-} \] \[ \tau^{+} + \tau^{-} = 0 \]

Example

Given the see-saw is balanced, find force \(P\).

Dog on seesaw

\[ \text{Taking moments about A} \] \[ \tau^{-} = 0.5 \times 10g + 2.5 \times 50g \] \[ = 130g \]

\[ \tau^{+} = 1 \times P \] \[ = P \]

\[ \tau^{+} = \tau^{-} \] \[ P = 130g \]

Static Rigid Bodies

The total moment about any fixed point of the body is zero.

\[ \sum_{i}\tau_{i}^{+} \;-\; \sum_{j}\tau_{j}^{-} = 0 \] \[ \sum \vec{F} = 0, \qquad \sum \tau = 0 \]

Procedure

Draw a force diagram.
Resolve forces horizontally and vertically.
Set total forces to zero.
Take moments about a convenient point.
Classify clockwise/anticlockwise.
Set total moment to zero.

Example

A uniform beam, 3m long and of mass 20 Kg, is supported horizontally by two vertical ropes.

Each of the ropes is attached at a distance of 50 cm from the edge of the beam.

What weight in Newtons should be placed on one end of the beam

a) to remove the tension in one of the ropes ?
b) to make the tension in rope B double that of rope A?/li>

Beam diagram

a) Removing tension in rope A

Place a weight (W) on the far right hand edge of the beam:

Beam with weight

The immediate affect of W is to turn the beam clockwise, which will loosen the left hand rope. So when no tension, T₁ = 0

Taking moments about B:

\[ \tau^{+} = 1 \times 20g \] \[ = 20g \] \[ \tau^{-} = 2T_{1} + 0.5W \] \[ = 2 \times 0 + 0.5W \] \[ = 0.5W \]

\[ \sum_{i}\tau_{i}^{+} \;-\; \sum_{j}\tau_{j}^{-} = 0 \] \[ 20g - 0.5W = 0 \] \[ 20g = 0.5W \] \[ W = 40g \]

A weight of 40g N (approximately 392.4 N) is required.

b) Doubling tension in rope B

Again, place a weight (W) on the far right hand edge of the beam:

Beam diagram 2

Taking moments about the right hand edge:

\[ \sum_{i}\tau_{i}^{+} \;-\; \sum_{j}\tau_{j}^{-} = 0 \] \[ 20g - 0.5W = 0 \] \[ 20g = 0.5W \] \[ W = 40g \]

\[ \sum_{i}\tau_{i}^{+} \;-\; \sum_{j}\tau_{j}^{-} = 0 \] \[ 30g - 3.5T = 0 \] \[ 30g = 3.5T \] \[ T = \frac{30g}{3.5} = \frac{60g}{7} \]

Resolving forces vertically,

\[ T + 2T - 20g - W = 0 \]

\[ W = \frac{60g}{7} + \frac{120g}{7} - 20g \] \[ = \frac{180g}{7} - \frac{140g}{7} \] \[ = \frac{40g}{7} \]

A weight of 40g/7 N (approximately 56.06 N) is required.

Ladder Against a Wall

A uniform ladder of mass \(M\) and length \(l\) rests against a smooth wall. Ground is rough with \(\mu = 1\). Find the minimum angle \(\theta\) for equilibrium.

The center of gravity is the midpoint of the ladder.

S and R are the horizontal and vertical normal reaction of the ladder on the wall.

T is frictional force of the base of the ladder.

Ladder diagram

Taking moments about the bottom of the ladder :

\[ \tau^{+} = Mg \times \frac{l\cos\theta}{2} \] \[ = \frac12 Mgl\cos\theta \] \[ \tau^{-} = S\,l\sin\theta \] \[ \tau^{-} = S\,l\sin\theta \]

\[ \sum_{i}\tau_{i}^{+} \;-\; \sum_{j}\tau_{j}^{-} = 0 \] \[ \frac12 Mgl\cos\theta - Sl\sin\theta = 0 \] \[ S = \frac{Mgl\cos\theta}{2l\sin\theta} \] \[ S = \frac{Mg\cos\theta}{2\sin\theta} \] \[ S = \frac{Mg\cot\theta}{2} \] \[ S = \frac12 Mg\cot\theta \]

Resolving forces vertically:

\[ R = Mg \]

Resolving forces horizontally:

\[ T = S = \frac12 Mg\cot\theta \]

\[ T \le \mu R \] \[ \frac12 Mg\cot\theta \le \mu R \] \[ \mu \ge \frac{Mg\cot\theta}{2R} \] \[ \text{Remember } R = Mg \] \[ \mu \ge \frac{\cot\theta}{2} \] \[ 2\mu \ge \cot\theta \] \[ \cot\theta \le 2\mu \] \[ \theta \le \cot^{-1}(2\mu) \]

The inverse cotangent must lie between 0 and \( \pi \) rads.

Using

\[ \cot^{-1}(x) = \frac{\pi}{2} - \tan^{-1}(x) \]

\[ \cot^{-1} 2 = \frac{\pi}{2} - \tan^{-1}(2) \] \[ \cot^{-1} 2 = \frac{\pi}{2} - 1.1071487\ \text{rads} \] \[ = 0.463647609 \] \[ = 0.463647609 \times \frac{180}{\pi}\ \text{degrees} \] \[ = 26.57^\circ\ \text{(2 dp)} \]

Couples

Like forces are parallel and act in the same direction.

Unlike forces are parallel and act in opposite directions.

A couple consists of two equal and opposite parallel forces not acting along the same line.

Couple Couple diagram

Not a Couple like4

Example

Show that the forces on the rectangle form a couple and find its moment.

Rectangle forces

Resolving forces parallel to BA gives

\[ \overrightarrow{BA} + \overrightarrow{CD} \] \[ = -3N + 3N \] \[ = 0 \]

So BA and CD form a couple

Resolving forces parallel to BC gives

\[ \overrightarrow{BC} + \overrightarrow{AD} \] \[ = -5N + 5N \] \[ = 0 \]

So BC and AD form a couple

The system has no translatory effect, so must either be in equilibrium or reduce to a couple.

Taking moments anticlockwise about O

\[ \tau_{O}^{+} = (3 \times 3) + (-5 \times 2) + (3 \times 3) + (-5 \times 2) \] \[ = 18 - 20 \] \[ = -2 \] \[ = 2\,N\,m\ \text{clockwise} \]

Taking moments anticlockwise about A

\[ \tau_{A}^{+} = (3 \times 0) + (-5 \times 4) + (3 \times 6) + (-5 \times 0) \] \[ = -20 + 18 \] \[ = -2 \] \[ = 2\,N\,m\ \text{clockwise} \]

Taking moments anticlockwise about B

\[ \tau_{B}^{+} = (3 \times 0) + (-5 \times 0) + (3 \times 6) + (-5 \times 4) \] \[ = 18 - 20 \] \[ = -2 \] \[ = 2\,N\,m\ \text{clockwise} \]

Taking moments anticlockwise about C

\[ \tau_{c}^{+} = (3 \times 0) + (-5 \times 0) + (-5 \times 4) + (3 \times 6) \] \[ = -20 + 18 \] \[ = -2 \] \[ = 2\,N\,m\ \text{clockwise} \]

Taking moments anticlockwise about D

\[ \tau_{D}^{+} = (3 \times 0) + (-5 \times 0) + (3 \times 6) + (-5 \times 4) \] \[ = 18 - 20 \] \[ = -2 \] \[ = 2\,N\,m\ \text{clockwise} \]

The moment of the couple is 2 N·m clockwise, so a 2 N·m anticlockwise couple is required for equilibrium.