3.2.1 - Dynamics and free-body diagrams

Net Force and the Newton

A force is a vector. That means it has both a magnitude and a direction. When several forces act on an object, the force that matters for its acceleration is the resultant force, also called the net force.

If the resultant force on an object is zero, the object has zero acceleration. It may be at rest, or it may be moving with constant velocity. If the resultant force is not zero, the object accelerates in the direction of the resultant force.

In this equation, $F$ is the resultant force in newtons $\mathrm{N}$, $m$ is the mass in kilograms $\mathrm{kg}$, and $a$ is the acceleration in metres per second squared $\mathrm{m} \mathrm{s}^{-2}$. OCR expects you to recall this equation.

The unit newton follows directly from $F = ma$.

So:

Worked example: acceleration from a resultant force

A trolley of mass $2.4\,\mathrm{kg}$ has a horizontal pull of $18\,\mathrm{N}$ to the right and a frictional force of $6.0\,\mathrm{N}$ to the left.

Choose right as positive.

Resultant force:

$F = 18 - 6.0 = 12\,\mathrm{N}$

Apply $F = ma$:

The acceleration is $5.0\,\mathrm{m}\,\mathrm{s}^{-2}$ to the right.

Notice the wording: the pull is not automatically the resultant force. You first add the forces with directions, then use $F = ma$.

Weight and Common Forces

Mass and weight are not the same quantity. Mass is a measure of inertia and is measured in kilograms. Weight is a force caused by a gravitational field and is measured in newtons.

Here, $W$ is weight in newtons $\mathrm{N}$, $m$ is mass in kilograms $\mathrm{kg}$, and $g$ is gravitational field strength in $\mathrm{N} \mathrm{kg}^{-1}$. Near Earth's surface, $g$ is about $9.81\,\mathrm{N}\,\mathrm{kg}^{-1}$, which is equivalent to an acceleration of free fall of $9.81\,\mathrm{m}\,\mathrm{s}^{-2}$ in calculations. OCR expects you to recall $W = mg$.

Worked example: weight from mass

A student has mass $64\,\mathrm{kg}$. Estimate the student's weight near Earth's surface.

To a sensible number of significant figures, the weight is about $630\,\mathrm{N}$.

The named forces in this OCR section are:

• tension: a pulling force exerted by a string, rope, cable or wire along its length
• normal contact force: a contact force perpendicular to the surface touching the object
• upthrust: an upward force exerted by a fluid on an object in the fluid
• friction: a contact force parallel to a surface that opposes relative motion or the tendency to slide

These names describe forces acting on the chosen object. For example, if a box is pulled by a rope, the rope exerts tension on the box. If the box rests on a floor, the floor exerts a normal contact force on the box.

A common error is to assume the normal contact force is always equal to weight. It is only equal to weight when the vertical forces balance in that particular situation. If there are extra vertical forces, or if the object accelerates vertically, the normal contact force may be different from $mg$.

Free-Body Diagrams

A free-body diagram is a force model for one selected object. It strips away the surroundings and shows only the forces acting on that object.

[DIAGRAM: asset_name: 3.2.1 - Dynamics and free-body diagrams - diagram 1; asset_slug: 3_02_1_dynamics_and_free_body_diagrams__diagram_01; recommended_method: drawn_physics; caption: Free-body diagrams show the forces acting on each chosen object.; description: Free-body diagrams for three chosen objects distinguish weight, contact forces, drag and upthrust.]

Use this routine whenever a force question feels crowded.

1. Choose the object or system.
2. Replace it with a dot or simple shape.
3. Add only forces acting on that object or system.
4. Label each force with a name, symbol and direction where possible.
5. Choose axes or a positive direction.
6. Add forces along each axis to find the resultant force.

Do not add forces that the object exerts on something else. For example, the weight of a book is the gravitational force of Earth on the book. The book also pulls on Earth gravitationally, but that force acts on Earth, so it does not belong on the book's free-body diagram.

Do not use a velocity arrow as if it were a force. A falling object can be moving downward while its resultant force is upward, in which case it is accelerating upward and slowing down. Velocity direction and acceleration direction are separate ideas.

Worked example: reading a free-body diagram in words

An object is moving downward through water. Its weight is $1.20\,\mathrm{N}$ downward. Upthrust is $0.80\,\mathrm{N}$ upward and drag is $0.60\,\mathrm{N}$ upward.

Take upward as positive.

The resultant force is $0.20\,\mathrm{N}$ upward. Therefore the acceleration is upward. Since the object is moving downward but accelerating upward, it is slowing down.

Constant Force in One Dimension

When the resultant force on a constant-mass object is constant, its acceleration is constant. In one dimension, choose a positive direction and keep signs consistent.

The calculation route is:

• choose the object or system
• choose a positive direction
• add forces in that direction, subtract forces in the opposite direction
• use $F = ma$
• interpret the sign of $a$

Worked example: upward lift acceleration

A lift of mass $520\,\mathrm{kg}$ has tension $5600\,\mathrm{N}$ upward in the cable. Its weight is $5100\,\mathrm{N}$ downward. Find the acceleration.

Choose upward as positive.

Resultant force:

$F = 5600 - 5100 = 500\,\mathrm{N}$

Apply $F = ma$:

The acceleration is $0.96\,\mathrm{m}\,\mathrm{s}^{-2}$ upward.

The lift could be moving upward and speeding up, moving downward and slowing down, or instantaneously at rest and about to move upward. The force calculation tells you acceleration, not the full motion history.

For a connected system, decide whether the forces you are using are external forces on the whole accelerating system. A trolley pulled by a hanging mass is a good example. If the trolley and hanging mass accelerate together and friction is negligible, the driving force is the hanging weight, but the mass being accelerated is the trolley plus the hanging mass.

Assume a light inextensible string and a frictionless pulley with negligible rotational inertia: the masses share the same acceleration magnitude and the string has the same tension on each side.

Suppose a $1.2\,\mathrm{kg}$ trolley is pulled by a hanging $0.30\,\mathrm{kg}$ mass. The weight of the hanging mass is:

The total mass being accelerated is:

$1.2 + 0.30 = 1.5\,\mathrm{kg}$

For the whole system, neglecting friction:

Using only the trolley mass would overestimate the acceleration because the hanging mass also has to be accelerated.

Constant Force in Two Dimensions

Two-dimensional dynamics uses the same idea as one-dimensional dynamics, but the forces must be added as vectors. Choose perpendicular axes, usually horizontal and vertical, and apply $F = ma$ separately along each axis.

Here, $\sum F_{x}$ is the resultant force along the x-axis and $\sum F_{y}$ is the resultant force along the y-axis. The accelerations $a_{x}$ and $a_{y}$ are the components of the acceleration.

[DIAGRAM: asset_name: 3.2.1 - Dynamics and free-body diagrams - diagram 2; asset_slug: 3_02_1_dynamics_and_free_body_diagrams__diagram_02; recommended_method: drawn_physics; caption: Perpendicular force components determine the resultant force and acceleration.; description: A 3-4-5 force-component triangle determines the resultant and an acceleration of 10 m per second squared.]

If the resultant force is constant, both acceleration components are constant. That is why constant-force problems can connect naturally to constant-acceleration kinematics: once you know $a_{x}$ and $a_{y}$, you can use the usual constant-acceleration relationships along each axis.

Worked example: perpendicular forces

A small object of mass $0.50\,\mathrm{kg}$ experiences a resultant force of $3.0\,\mathrm{N}$ east and $4.0\,\mathrm{N}$ north.

Acceleration components:

$a_{x} = F_{x} / m = 3.0 / 0.50 = 6.0\,\mathrm{m}\,\mathrm{s}^{-2}$ east

$a_{y} = F_{y} / m = 4.0 / 0.50 = 8.0\,\mathrm{m}\,\mathrm{s}^{-2}$ north

Magnitude of acceleration:

Direction:

$\tan \theta = 8.0 / 6.0$, so $\theta = 53 {}^{\circ}$ north of east.

You could also first find the resultant force magnitude:

Then:

Both routes agree because force and acceleration are parallel vectors for a constant mass.

From rest, the same constant component accelerations $a_x=6.0\,\mathrm{m\,s^{-2}}$ east and $a_y=8.0\,\mathrm{m\,s^{-2}}$ north act for $2.0\,\mathrm{s}$. Apply constant-acceleration equations independently: $v_x=a_xt=12\,\mathrm{m\,s^{-1}}$, $v_y=a_yt=16\,\mathrm{m\,s^{-1}}$, giving speed $20\,\mathrm{m\,s^{-1}}$ at $53^\circ$ north of east. The displacement components are $s_x=\frac12a_xt^2=12\,\mathrm{m}$ and $s_y=\frac12a_yt^2=16\,\mathrm{m}$, giving displacement $20\,\mathrm{m}$ in the same direction.

Force–acceleration data

Although 3.2.1 has no explicit PAG reference, OCR can still assess dynamics through practical-style data. A common route is to vary the resultant force on a trolley and measure its acceleration.

From $F = ma$, if mass is constant:

$F = m a$

A graph of resultant force $F$ on the y-axis against acceleration $a$ on the x-axis should be a straight line. The gradient is:

$\text{gradient}=\Delta F/\Delta a$

Since $F / a = m$, the gradient has units:

$N / (m s^{-2}) = kg$

So the gradient gives the mass being accelerated. For true resultant force, the line passes through the origin. A small positive measured intercept means the plotted force overestimates that resultant, for example because of a force-sensor zero offset or friction not yet subtracted from the measured pulling force.

Worked example: graph gradient

A best-fit line on a graph of $F$ against $a$ passes through the points:

• $a = 0.50\,\mathrm{m}\,\mathrm{s}^{-2}$, $F = 0.40\,\mathrm{N}$
• $a = 2.50\,\mathrm{m}\,\mathrm{s}^{-2}$, $F = 2.00\,\mathrm{N}$

Gradient:

The mass found from the graph is $0.80\,\mathrm{kg}$.

In a trolley-and-hanger experiment, vary the driving force by transferring masses from the trolley to the hanger. This keeps the total accelerating mass constant. Measure acceleration with light gates or motion tracking, repeat each setting and estimate friction independently. Plot the actual resultant force after subtracting opposing forces. Plotting the hanging weight without this correction gives a positive force intercept if friction is approximately constant.

[DIAGRAM: asset_name: 3.2.1 - Dynamics and free-body diagrams — figure 3; asset_slug: 3_02_1_dynamics_and_free_body_diagrams__diagram_03; recommended_method: drawn_physics; caption: The force-acceleration gradient gives mass while the intercept represents a constant force offset.; description: An applied-force-against-acceleration line has a 0.20 N intercept and a gradient equal to the 0.80 kg mass.]