Free-body diagrams and Newton's lawsEdexcel International A Level Physics: Revision notes

A free-body diagram shows a single object (a particle or an extended rigid body) as a dot or outline, with an arrow for every force acting on it, labelled by name and drawn from the point where it acts. Forces exerted by the object on other things are not drawn.

Typical forces: weight (down), normal contact force (perpendicular to a surface), tension, driving force, friction, drag or air resistance (opposing motion).

The centre of gravity of an extended body is the point through which the whole weight appears to act. For a uniform body it is at the geometric centre. Draw the weight arrow from this point.

Newton's first law: an object remains at rest or moves at constant velocity unless a resultant force acts on it. When the forces are balanced, $a = 0$.

Newton's second law (constant mass): the resultant force is proportional to the rate of change of momentum, giving $\Sigma F = ma$, where $\Sigma F$ is the resultant force in N, $m$ the mass in kg and $a$ the acceleration in m s⁻² in the direction of the resultant force.

Example: a van of mass 1500 kg with driving force 3200 N and resistive force 800 N has $a = \frac{3200 - 800}{1500} = 1.6\ \text{m s}^{-2}$.

In problems, resolve the forces along the direction of motion, find the resultant, then apply $F = ma$.

The gravitational field strength $g$ is the force per unit mass on a small mass in a gravitational field, $g = \frac{F}{m}$, in N kg⁻¹. Near the Earth's surface $g = 9.81\ \text{N kg}^{-1}$, which is also the free-fall acceleration in m s⁻².

Weight is the force of gravity on an object, $W = mg$. Mass is a scalar measured in kg; weight is a vector measured in N acting downwards.

In free fall with no air resistance, $a = \frac{mg}{m} = g$, which is why all objects fall with the same acceleration regardless of mass.

Example: a 70 kg person in a lift accelerating up at $1.2\ \text{m s}^{-2}$: $R - mg = ma$, so $R = 70(9.81 + 1.2) = 771\ \text{N}$.

When an object falls through a fluid, drag (air or fluid resistance) increases with speed.

1. At release the speed is zero, so there is no drag and $a = g$
2. As the speed increases, drag increases, so the resultant force $W - R$ falls and $a$ falls
3. At terminal velocity the drag equals the weight, the resultant force is zero and $a = 0$, so the object moves at constant velocity (Newton's first law)

The same idea applies to a vehicle whose driving force equals the resistive forces at its top speed.

Release a small, dense steel ball from an electromagnet at a measured height $h$ above a trapdoor connected to an electronic timer. The timer starts when the current is switched off and stops when the ball opens the trapdoor.

• Measure $h$ with a metre rule; record the time $t$
• Repeat each height and take a mean time, then use a range of heights
• Since $h = \frac{1}{2}gt^2$, plot $h$ against $t^2$ and find $g = 2 \times$ gradient

Sources of error: air resistance (use a small dense ball), delayed release due to residual magnetism (a thin non-magnetic spacer), and the uncertainty in measuring $h$.

Newton's third law: when body A exerts a force on body B, body B exerts a force on body A that is equal in magnitude, opposite in direction, of the same type and acts on the other body at the same time.

The forces of a pair always act on different bodies. The weight of a book (Earth pulling the book) and the contact force from the table (table pushing the book) are not a pair: both act on the book and are different types. The pair of the weight is the gravitational pull of the book on the Earth.