Dynamics and Forces

Diagram-to-equation workflow

01 / Diagram-to-equation workflow

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Use the same problem-solving order every time
Step	What to put on paper
1. Choose the system	Name the one object or system being analyzed
2. Draw the free-body diagram	Include every external force acting on that system—and no motion arrows
3. Choose axes	State which directions are positive
4. Write dynamics by axis	Use x = max and y = may separately
5. Substitute and solve	Keep signs and units attached
6. Interpret	Translate the sign of the result into a physical direction

Choose the system, draw only the external forces, choose axes, write one Newton’s second law equation for each axis, solve with signs and units, and interpret the result.

02 / Thrown-ball forces

A ball can move upward while the net force points downward

Consider a ball after it has left the hand but before it reaches its highest point. The ball’s velocity points upward. If air resistance is neglected, the only force acting on the ball is its weight, directed downward.

The hand is no longer touching the ball, so there is no continuing upward applied force. The downward net force produces a downward acceleration even while the ball is still moving upward.

Animation of a marker moving vertically upward, reaching a highest point, and returning downward along a vertical axis. — The motion reverses at the highest point, but the gravitational acceleration remains downward throughout. This animation shows position, not a free-body diagram. Animation by Yuta Aoki, CC BY-SA 3.0.

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The same ideal free-body diagram applies at three different moments
Moment	Velocity	Force and acceleration
Rising	Upward	Weight downward; acceleration downward
At the highest point	Zero for one instant	Weight downward; acceleration downward
Falling	Downward	Weight downward; acceleration downward

Ignoring air resistance, the ball has only a downward weight force and downward acceleration while rising, at the highest point, and while falling; only its velocity changes.

Check your understanding

At the highest point, what are the ball’s velocity, acceleration, and net force if air resistance is neglected?

Show the answer

The instantaneous velocity is zero. Acceleration points downward with magnitude g, and the net force is the ball’s weight, mg, downward.

Zero velocity at one instant does not imply zero acceleration or zero net force.

Interactive practice

Use the Free-Body Diagram Builder and Checker to choose the real forces, set their directions, and check the completed diagram before moving to equations.

Starting interactive…

Open Free-Body Diagram Builder and Checker on its own page

03 / Accelerating car

An accelerating car requires two separate force equations

Now consider a car accelerating to the right on a level road. In the simplified model used here, the forward driving force is larger than the leftward resistive force. Vertically, weight acts downward and the road’s normal force acts upward.

The car has no vertical acceleration, so the vertical forces balance. The horizontal forces do not balance, so their signed sum equals max.

Three-part OpenStax figure showing people pushing a car, the corresponding free-body diagram with forward forces, friction, normal force, and weight, and a tow truck producing a larger forward force and acceleration. — A car’s free-body diagram includes the external forces acting on the chosen car system. The acceleration and net-force arrows shown outside the force inventory describe the result; they are not additional forces. OpenStax University Physics Volume 1, Figure 5.10, CC BY 4.0.

Write one Newton’s second law equation for each axis

ᵧ = F(normal) − F(weight) = 0; ₓ = F(drive) − F(resistive) = maₓ

FN: normal force exerted by the road on the car (newton (N))
Fg: weight of the car (newton (N))
Fdrive: forward external driving force in this model (newton (N))
Fresist: leftward resistive force in this model (newton (N))
ax: horizontal acceleration (meter per second squared (m/s²))

Assumptions

Choose right and up as positive.
The road is level and the car has no vertical acceleration.
The listed forces are the complete external-force inventory for the simplified model.

04 / Complete force example

Solve a complete two-axis force problem

A 12 kg object is supported on a horizontal surface. It experiences 8 N right, 7 N left, and 10 N left. Find its weight, normal force, and horizontal acceleration. Use g = 9.81 m/s².

Assumptions

Choose right and up as positive.
The object has no vertical acceleration.
The stated forces are the complete force inventory.

Find the weight magnitude
F(weight) = mg = (12 kg)(9.81 m/s²) = 117.72 N
Weight acts downward. The calculated value is its magnitude.
Use vertical equilibrium
ᵧ = F(normal) − F(weight) = 0 → F(normal) = 117.72 N
In this specific situation, the normal force equals the weight because no other vertical forces act and ay = 0.
Add the horizontal forces with signs
ₓ = 8 N − 7 N − 10 N = −9 N
The negative net force points left.
Apply Newton’s second law
aₓ = ₓ/m = (−9 N)/(12 kg) = −0.75 m/s²
The acceleration points left. The sign does not reveal the object’s current velocity.

Result: Weight is 117.72 N downward, normal force is 117.72 N upward, and acceleration is 0.75 m/s² left.

Check your understanding

The calculated acceleration is −0.75 m/s². Must the object be moving left?

Show the answer

No. The negative sign says the acceleration points left under the chosen axis convention. The object could be moving left, at rest at that instant, or moving right while slowing down.

Acceleration describes the direction of velocity change, not necessarily the current direction of velocity.

Interactive practice

Use the Physics Problem Workbench for generated one-dimensional Newton’s second law problems. Keep the knowns visible, choose the equation from the signed force model, isolate the unknown, substitute with units, and check the physical meaning of the result.

Starting interactive…

Open Physics Problem Workbench on its own page

05 / Angled push

An angled push changes both horizontal and vertical force balances

When a lawnmower is pushed down and to the right along its handle, the applied force has a rightward component and a downward component. The rightward component can balance friction during constant-velocity motion.

The downward component adds to the mower’s weight. The ground must therefore exert a normal force larger than the weight alone. This is why FN = Fg is not a universal rule.

Constant-velocity lawnmower model

ₓ = F(applied,x) − f = 0; ᵧ = F(normal) − F(weight) − F(applied,y) = 0

Fapp,x: rightward component of the applied push (newton (N))
Fapp,y: downward component magnitude of the applied push (newton (N))
f: friction opposing the mower’s sliding motion (newton (N))
FN: normal force exerted by the ground (newton (N))

Assumptions

The mower moves at constant velocity, so acceleration is zero in both axes.
The applied push is resolved into perpendicular horizontal and vertical components.
The applied vertical component points downward.

Check your understanding

For the constant-velocity lawnmower, is the normal force smaller than, equal to, or larger than the mower’s weight?

Show the answer

It is larger: FN = Fg + Fapp,y. The surface supports both the mower’s weight and the downward component of the applied push.

Normal force must be determined from the vertical dynamics equation; it is not automatically equal to weight.

06 / AP Physics focus

AP Physics Focus: model first, then calculate

AP Physics 1 problems reward a defensible representation: identify the system, draw only external forces, choose axes, and write separate component equations before substituting numbers. A velocity or acceleration arrow may be shown near a free-body diagram for context, but it must not be counted as a force.

This page preserves the lesson’s 9.81 m/s² value in the numerical example. On AP Physics 1 work, use the value supplied by the problem; when none is supplied, AP allows and encourages the convenient near-Earth approximation g = 10 m/s².

Check your understanding

A student writes FN = Fg for every object on a horizontal surface. Use the lawnmower example to evaluate the claim.

Show the answer

The claim is false. FN equals Fg only when the vertical force sum and acceleration make that relationship true. A downward applied component gives FN = Fg + Fapp,y.

Normal force is a contact response determined from the full vertical force equation.

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What each example is meant to teach
Example	Central decision
Ball after release	Motion direction does not create an extra force
Accelerating car	Separate balanced vertical forces from unbalanced horizontal forces
12 kg object	Add signed forces first, then divide the net force by mass
Angled lawnmower push	Resolve the applied force and do not assume FN = Fg

The examples distinguish motion from force, separate axes, use signed net force in Newton’s second law, and show why an angled applied force changes the normal force.