3.2MechanicsFoundation

Forces and Free-Body Diagrams

Before Newton’s laws can be applied to anything, you first need to see clearly what is actually pushing and pulling on an object. A free-body diagram strips away every distraction and shows exactly that — and resolving forces on a slope is the classic test of whether you can really use one.

35 min Video by Physics Online Module 3.2 Forces in action

Free Body Diagrams and Objects on an Inclined Plane — A Level PhysicsWatch the full walkthrough before the notes below.

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What you'll be able to do

Name the common forces: weight, normal contact force, tension, friction, drag and thrust
Draw a free-body diagram for an object in a given situation
Resolve forces on an object on an inclined plane
Apply the condition for equilibrium to a system of forces

Common forces

Every force in this module falls into a small set of familiar types: $weight$ (the pull of gravity, always vertically downward from the centre of mass), the $normal contact force$ (perpendicular to whatever surface an object rests on), $tension$ (in a string or rope), $friction$ and $drag$ (both opposing relative motion), and driving forces such as $thrust$ .

W = m g

Weight (N) = mass (kg) × gravitational field strength (N/kg).

Free-body diagrams

A $free-body diagram$ isolates a single chosen object and shows only the forces acting $on$ it — never the forces it exerts on something else. Each force is drawn as an arrow starting at the object, with length suggesting relative size and direction showing which way it acts.

Tip — A simple check: if you can’t say in one word what is causing a force ("gravity", "the rope", "the floor"), it probably shouldn’t be on the diagram.

Objects on an inclined plane

For an object resting on (or sliding along) a slope, it is almost always easiest to resolve forces along the slope and perpendicular to it, rather than horizontally and vertically. Weight, the one force that doesn’t naturally align with the slope, must be split into a component acting down the slope ( $m g sin θ$ ) and a component pressing into the slope ( $m g cos θ$ ).

W_{∥} = m g sin θ W_{⊥} = m g cos θ

Weight resolved parallel and perpendicular to a slope inclined at angle θ.

1Weight:

W = m g = 4.0 \times 9.81 = 39.24

2Component down the slope:

W_{∥} = W sin 25° = 39.24 \times 0.4226

3Normal contact force (equal and opposite to the perpendicular weight component, since there’s no acceleration perpendicular to the slope):

N = W cos 25° = 39.24 \times 0.9063

Answer

W_{∥} \approx 16.6

N (down the slope);

N \approx 35.6

Tip — On a frictionless slope, the normal contact force is NOT simply equal to the object’s full weight — it only balances the PERPENDICULAR component, mg cos θ, which is smaller than mg for any nonzero angle.

Equilibrium

An object is in $equilibrium$ when the resultant of all forces acting on it is zero — it is either stationary or moving at constant velocity. For an object in equilibrium, the forces along any chosen direction (such as along a slope) must sum to zero, which is exactly the condition used to find an unknown force, such as friction or tension, holding the system in balance.

1For equilibrium along the slope, friction must exactly balance the component of weight acting down the slope.

2Friction force

= W_{∥} \approx 16.6

N (up the slope).

AnswerFriction force ≈ 16.6 N, acting up the slope.

Equation recap

W = m g

Weight.

W_{∥} = m g sin θ, W_{⊥} = m g cos θ

Weight resolved on a slope.

Common mistakes to avoid

Adding a force the object exerts on something else to its own free-body diagram.

A free-body diagram shows only forces acting ON the chosen object, never forces it exerts elsewhere (Newton’s third law pairs act on different objects).

Assuming the normal contact force on a slope always equals the object’s full weight.

On an incline, the normal force only balances the PERPENDICULAR component of weight (mg cos θ) — it is smaller than mg whenever θ > 0.

Mixing up sinθ and cosθ when resolving weight on a slope.

The component ALONG the slope uses sinθ; the component perpendicular (into the slope) uses cosθ — sketch the triangle to check if unsure.

Key takeaways

Common forces: weight (W=mg), normal contact force, tension, friction, drag, thrust.
A free-body diagram shows only the forces acting on one chosen object.
On a slope, weight resolves into mg sinθ (down the slope) and mg cosθ (into the slope).
An object in equilibrium has zero resultant force — forces along any direction must balance.

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Forces and Free-Body Diagrams

35 min Video by Physics Online Module 3.2 Forces in action