Chapter 5: Force and Motion

1. What is a Force? Why does it matter?

A force is simply a push or a pull that one object (the agent) exerts on another.
Every force has magnitude (how strong) and direction → we treat forces as vectors.
Two broad families:
1. Contact forces – need physical touch (e.g., friction, tension).
2. Long-range forces – act at a distance (e.g., gravity, magnetism).

2. Net Force & Newton's Second Law

Net force = the vector sum of all forces acting on an object:
$\vec{F}_{\text{net}} = \sum_i \vec{F}_i$
Experiments show:
1. Acceleration $\vec{a}$ grows in direct proportion to $\vec{F}_{\text{net}}$ .
2. For a given $\vec{F}_{\text{net}}$ , heavier objects (more mass $m$ ) accelerate less.
Combining both ideas gives Newton's 2nd Law:
$\boxed{\vec{a} = \frac{\vec{F}_{\text{net}}}{m}}$
The acceleration points in the same direction as $\vec{F}_{\text{net}}$ .
Units:
- Mass → kilograms (kg)
- Force → newtons (N) – one newton is $1\ \text{kg} \cdot \text{m/s}^2$ .

3. Equilibrium & Newton's First Law

If no net force acts ( $\sum \vec{F} = 0$ ), the object is in equilibrium:
- At rest → it stays at rest.
- Moving at constant velocity → it keeps that velocity.
- Written as: $\vec{a} = 0$ .
This is Newton's 1st Law ("law of inertia").

Key takeaway: motion itself needs no force to continue—only a change in motion needs force.

4. Everyday Forces to Recognise

Symbol	Force name	Direction rule	Useful equation	Comment
$\vec{F}_G$	Gravity	Straight down (toward Earth's centre)	$F_G = mg$	Long-range
$\vec{F}_{\text{sp}}$	Spring	Toward spring's relaxed (equilibrium) length	—	Hooke's law appears later
$\vec{T}$	Tension	Along (and pulling) the string/rope	—	Same at every point in an ideal rope
$\vec{n}$	Normal	Perpendicular to the surface	—	"Support" force
$\vec{f}_k$	Kinetic friction	Parallel & opposite to sliding motion	$f_k = \mu_k n$	Surface must be sliding
$\vec{f}_s$	Static friction	Parallel & opposite to intended motion	$f_s \leq \mu_s n$	Holds an object at rest

Free-body diagram: draw the object alone, then add each of these forces with correct directions and labels. That picture is half the battle!

5. From Straight-Line Motion to Rotation – Introducing Torque

A force can also twist an object about a pivot.
We measure this twisting tendency with torque $\vec{\tau}$ :
$\boxed{\vec{\tau} = \vec{r} \times \vec{F}}$
where:
- $\vec{r}$ = position vector from the pivot to where the force is applied,
- $\times$ = vector (cross) product.
Magnitude:
$\tau = F\,r\sin\phi = F\,d_\perp$
- $\phi$ = angle between $\vec{F}$ and $\vec{r}$ .
- $d_\perp$ = shortest (perpendicular) distance from pivot to the force's line of action.
Sign convention (viewing +z coming out of the page):
- Counter-clockwise (CCW) twist → $+\tau$
- Clockwise (CW) twist → $-\tau$
Component trick: sometimes it is faster to resolve $\vec{F}$ into $x$ - and $y$ -parts and add their torques:
$\tau = F_x\,d_{y,\perp} + F_y\,d_{x,\perp}$

6. Net Torque

Just like forces, torques add:

\vec{\tau}_{\text{net}} = \sum_i \vec{\tau}_i

The net torque tells you the overall rotational effect of all forces on the object.

7. Quick Practice (try before peeking!)

A. The square plate shown on the left has a pivot point O at its centre. If $F_1 = 26\ \text{N}$ , $F_2 = 14\ \text{N}$ , and $F_3 = 18\ \text{N}$ , what is the net torque?

Square plate with forces

Using the torque formula $\tau = Fr\sin\phi$ for each force:

For $F_1$ (26 N):

Distance from pivot to force application point: $r = 0.127\ \text{m}$
Angle between force and position vector: $\phi = 135^\circ$
Torque: $\tau_1 = -(26\ \text{N})\sin135^\circ(0.127\ \text{m}) = -2.33\ \text{Nm}$ (CW)

For $F_2$ (14 N):

Distance from pivot to force application point: $r = 0.127\ \text{m}$
Angle between force and position vector: $\phi = 135^\circ$
Torque: $\tau_2 = (14\ \text{N})\sin135^\circ(0.127\ \text{m}) = 1.26\ \text{Nm}$ (CCW)

For $F_3$ (18 N):

Distance from pivot to force application point: $r = 0.127\ \text{m}$
Angle between force and position vector: $\phi = 90^\circ$
Torque: $\tau_3 = (18\ \text{N})\sin90^\circ(0.127\ \text{m}) = 2.29\ \text{Nm}$ (CCW)

Net torque: $\tau_{\text{net}} = \tau_1 + \tau_2 + \tau_3 = -2.33\ \text{Nm} + 1.26\ \text{Nm} + 2.29\ \text{Nm} = 1.22\ \text{Nm}$ (CCW)

Click to reveal

B. The pulley shown on the right has a mass of $5.0\ \text{kg}$ and a diameter of $30\ \text{cm}$ . The pulley has a pivot point O located half way between its center and its rim. A rope holding two $15\ \text{kg}$ masses wraps around the pulley. What is the net torque?

Pulley with masses

We need to calculate the torque from the pulley's weight and the two hanging masses:

For the $15\ \text{kg}$ mass on the left:

Force: $F = mg = 15\ \text{kg} \times 9.8\ \text{m/s}^2 = 147\ \text{N}$
Distance from pivot to force line of action: $r = 0.075\ \text{m}$
Angle between force and moment arm: $\phi = 90^\circ$
Torque: $\tau_1 = (147\ \text{N})(0.075\ \text{m})\sin90^\circ = 11.025\ \text{Nm}$ (CCW)

For the pulley's weight ( $5.0\ \text{kg}$ ):

Force: $F = mg = 5.0\ \text{kg} \times 9.8\ \text{m/s}^2 = 49\ \text{N}$
Distance from pivot to center of mass: $r = 0.075\ \text{m}$
Angle between force and moment arm: $\phi = 90^\circ$
Torque: $\tau_2 = -(49\ \text{N})(0.075\ \text{m})\sin90^\circ = -3.675\ \text{Nm}$ (CW)

For the $15\ \text{kg}$ mass on the right:

Force: $F = mg = 15\ \text{kg} \times 9.8\ \text{m/s}^2 = 147\ \text{N}$
Distance from pivot to force line of action: $r = 0.225\ \text{m}$
Angle between force and moment arm: $\phi = 90^\circ$
Torque: $\tau_3 = -(147\ \text{N})(0.225\ \text{m})\sin90^\circ = -33.075\ \text{Nm}$ (CW)

Net torque: $\tau_{\text{net}} = \tau_1 + \tau_2 + \tau_3 = 11.025\ \text{Nm} + (-3.675\ \text{Nm}) + (-33.075\ \text{Nm}) = -25.725\ \text{Nm}$ (CW)