Forces

A force is a push or pull that can change motion. More precisely, the sum of all forces acting on an object determines its acceleration:

F=ma.\sum \vec{F} = m\vec{a}.

Newt the physics frog will help us build the idea from the ground up: first with no forces at all, then with spring-like forces and weight, and finally with the everyday contact forces that show up in free-body diagrams.

If vector arrows or component addition feel rusty, start with the Vectors page first. Free-body diagrams depend on adding force vectors, not just listing force names.

Force-Free Motion

If no force acts on an object, its velocity does not need to be zero. Zero acceleration simply means the velocity remains as it was. A short push can change Newt’s velocity, but otherwise Newt moves at constant velocity until the next force is applied.

F=0  v=constant.\sum \vec{F} = 0 \ \Rightarrow \ \vec{v} = \text{constant}.

Force-Free Newt

Tap anywhere to give Newt a short knock.

Springs

Springs are a clean place to start because their force depends on displacement. In the ideal model, the force points back toward the relaxed position:

Fs=kx.\vec{F}_s = -k\vec{x}.

The constant kk, called the spring constant, measures the stiffness. A larger displacement or a stiffer spring produces a larger restoring force.

Spring-Like Force

Drag Newt, then release. The spring force always redirects Newt's motion toward the relaxed position.

relaxedspring forcex = 85 px

This spring picture will become our microscopic model for contact forces. Atoms in a solid do not pass freely through one another. Push them a little too close and electric interactions push back; pull neighboring bits slightly apart and they tug back together. For small distortions, that push or pull often behaves like a tiny spring.

Gravity

Gravity does not require contact. Near Earth’s surface, we usually model it as a steady downward force:

Fg=mg.\vec{F}_g = m\vec{g}.

The magnitude of this force mgmg is known as the weight of the object. In the following diagrams, the force of gravity is then drawn as a weight arrow.

Gravity As a Downward Force

Drag Newt and release.

weight

Normal Force

The normal force is the contact force perpendicular to a surface. A table does not decide to push upward as a separate magic rule. Newt compresses the surface very slightly, and the surface’s microscopic spring-like structure pushes back.

Normal Force From Microscopic Springs

Drag Newt around the springy room, then release. Each wall compresses locally and pushes perpendicular to its surface.

weight

On a flat surface with no vertical acceleration, the normal force usually balances weight:

N=mg.N = mg.

On a ramp or in an accelerating system, the normal force may be smaller or larger than mgmg. What stays true is the direction: the normal force is perpendicular to the contact surface.

Tension

Tension can also be modeled as a spring-like contact force, but now the microscopic pieces are stretched instead of compressed. A rope can pull because neighboring bits of rope are tugged slightly apart, and those internal bonds pull back toward their relaxed spacing.

Newt has an unusually useful rope: his tongue.

Tension Through Newt's Tongue

Drag Newt and release. His tongue swings like a rope made from tiny stretched spring-like segments.

anchortensionweight
stretch
20 px
state
taut

A slack rope, string, cable, or tongue does not apply a tension force. Once it becomes taut, it can pull only along its own length. That is why tension arrows in free-body diagrams point away from the object along the rope.

Problem Solving

Weight and Normal Force

Problem

A 12 kg12\ \text{kg} box rests on a level floor with no vertical acceleration. Find the weight and the normal force magnitude. Use g=9.8 m/s2g = 9.8\ \text{m/s}^2.

Solve

The weight is

Fg=mg=(12)(9.8)=117.6 N.F_g = mg = (12)(9.8) = 117.6\ \text{N}.

With no vertical acceleration, the upward normal force balances the downward weight:

N=117.6 N.N = 117.6\ \text{N}.

Spring Force Direction

Problem

A spring with k=150 N/mk = 150\ \text{N/m} is stretched 0.08 m0.08\ \text{m} to the right. Find the spring force.

Solve

Hooke’s law gives

Fs=kx=(150)(0.08)=12 N.F_s = -kx = -(150)(0.08) = -12\ \text{N}.

The negative sign means the spring force points left, back toward equilibrium.

Tension in a Hanging Rope

Problem

A 5.0 kg5.0\ \text{kg} mass hangs at rest from a vertical rope. What is the rope tension?

Solve

At rest, the net force is zero. The upward tension balances the downward weight:

T=mg=(5.0)(9.8)=49 N.T = mg = (5.0)(9.8) = 49\ \text{N}.

Forces Checkpoint

Question 1 of 4

If the net force on an object is zero, what must be true about its motion?

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