Physics Library
 An open source physics library
Encyclopedia | Forums | Docs | Random |  
Login
create new user
Username:
Password:
forget your password?
Main Menu
Sections

Meta

Talkback

Downloads

Information
Common Forces in Mechanics (Topic)

Common Forces in Mechanics

Newton’s second law does not determine the forces by itself. It tells how the net force changes motion:

∑
    F = ma
(1)

for a constant-mass particle. To use this equation, the physical interactions acting on the chosen body must first be identified and modeled.

This article is a map of the force laws that occur most often in introductory mechanics. Each force has three questions attached to it:

  1. What physical interaction produces the force?
  2. In what direction does the force act?
  3. What model determines its magnitude?

The answers are not interchangeable. A normal force, for example, is recognized from a surface contact and points perpendicular to that surface, but its magnitude is usually found from Newton’s second law rather than from a universal formula.

PIC

Figure 1. Common force models arise from identifiable interactions between the chosen system and its environment. The force belongs on the free-body diagram only when that interaction crosses the system boundary.

1 Force laws are models

A force law is a mathematical model for an interaction. Some are fundamental or long-range, such as Newtonian gravity. Others are effective macroscopic descriptions of many microscopic interactions, such as a normal force, friction, drag, or an elastic spring force.

The same object may experience several forces at once. The procedure is therefore

                                                                   ∑
identify interactions    −→    write individual force vectors  − →        F = ma.
(2)

A label such as “applied force” describes how an external agent acts on the body, but it is not a universal force law with one fixed magnitude formula.

2 Weight

Near Earth’s surface, when The Gravitational Field can be treated as uniform, the gravitational force on a body of mass m is

|----------|
-W--=--mg.-|
(3)

Its magnitude is

W  = mg,
(4)

and its direction is vertically downward toward Earth.

The mass m is an intrinsic property of the object in Newtonian mechanics, while weight is a force and therefore depends on the local gravitational field. A body can have the same mass in different locations but different weights.

For problems extending over large distances, the uniform-field approximation may fail and Newton’s inverse-square gravitational law must be used instead.

3 Normal force

When two surfaces press against one another, the contact interaction generally has a component perpendicular to the surface. This perpendicular component is called the normal force.

Its direction is fixed geometrically:

The normal force acts perpendicular to the contact surface and pushes the bodies apart.

Its magnitude is not generally

N  = mg.
(5)

That equality occurs only in particular situations. The correct value of N normally follows from the force equation in the direction normal to the contact.

For example, if a block on a horizontal floor has no vertical acceleration and no other vertical forces, then

N  − mg  = 0,
(6)

so N = mg. If a rope pulls upward at an angle, or the supporting surface accelerates, the result changes.

4 Friction

Friction is the component of a contact force tangent to the interface. Its direction opposes the actual or impending relative sliding of the surfaces.

For static contact, the simple Coulomb model is

|--------------|
-0 ≤-fs ≤-μsN.--
(7)

The static friction force adjusts to the value required by the dynamics, up to the limiting magnitude

fs,max = μsN.
(8)

For surfaces sliding relative to one another, the introductory kinetic-friction model is

|----------|
-fk ≈-μkN.--
(9)

The kinetic-friction vector points opposite the relative sliding direction at the contact.

The coefficients μs and μk are empirical parameters. The simple Coulomb model is useful but not universal.

PIC

Figure 2. Surface forces are resolved into a normal component perpendicular to the contact and a friction component tangent to it. Their magnitudes must be determined from the appropriate contact model and the equations of motion.

5 Tension

A taut string, rope, or cable pulls on the body attached to it. The force is directed along the string, away from the body being analyzed.

For an ideal massless string passing over an ideal massless frictionless pulley, the Tension magnitude is the same throughout one continuous string:

T1 =  T2 = T.
(10)

This equality is a model assumption, not a universal property of every rope or pulley system. A massive rope can have different tension at different points because portions of the rope themselves require net force to accelerate.

6 Spring force

For an ideal linear spring, Hooke’s law is

|----------|
Fs =  − kx,|
------------
(11)

where x is the displacement of the spring endpoint from its equilibrium position along the spring axis and k is the spring constant.

In one dimension,

Fs = − kx.
(12)

The minus sign expresses the restoring nature of the force: an extension produces a force toward shorter length, while a compression produces a force toward longer length.

Hooke’s law is a linear approximation valid over the elastic range of the spring.

PIC

Figure 3. Tension acts along a taut string and pulls away from the body. An ideal linear spring exerts a restoring force along the spring axis toward its equilibrium configuration.

7 Drag

A body moving relative to a fluid experiences a resistive force called drag. The important velocity is the velocity relative to the fluid,

vrel = vbody − vfluid.
(13)

A common low-speed linear model is

|-------------|
FD  = − bvrel, |
--------------
(14)

where b is a positive drag coefficient.

A common quadratic model is

|------------------|
|F   = − c|v   |v   ,|
---D--------rel--rel-
(15)

with c > 0. In aerodynamic notation, the magnitude is often written

      1        2
FD  = --ρCDAv  rel.
      2
(16)

The direction of drag is opposite the relative velocity through the fluid. The correct drag model depends on the flow regime, body geometry, and Reynolds number.

8 Buoyant force

A body immersed in a fluid experiences a net pressure force that, in a hydrostatic fluid, gives the buoyant force

|--------------|
|FB = ρf Vdispg,|
----------------
(17)

where ρf is the fluid density and V disp is the displaced fluid volume.

The buoyant force points opposite the local effective direction of gravity; in the usual terrestrial setting it points upward.

Although buoyancy belongs mainly to fluid mechanics, it is useful to recognize it as another common external force in particle Free-body diagrams.

PIC

Figure 4. A fluid can exert both drag, which opposes relative motion through the fluid, and buoyancy, which results from the pressure field in the fluid.

9 Applied and contact forces

Introductory problems often specify an “applied force” directly, such as a person pushing a crate with 40 N. In that case the magnitude is given by the problem rather than by a separate constitutive law.

Likewise, the contact force between two bodies may be left as an unknown vector or separated into normal and tangential components. Newton’s third law then gives the corresponding force on the other body:

F      = − F     .
  A←B       B ←A
(18)

These two forces act on different bodies and therefore should not both appear on one body’s free-body diagram.

10 A compact force-recognition table

Interaction Typical direction Introductory magnitude model



Gravity near Earthdownward mg
Surface normal perpendicular to surfacesolve from dynamics
Static friction tangent to surface 0 ≤ fs ≤ μsN
Kinetic friction opposite sliding fk ≈ μkN
Tension along taut string solve from dynamics
Linear spring toward equilibrium Fs = k|x|
Linear drag opposite vrel FD = bvrel
Quadratic drag opposite vrel FD = cvrel2
Buoyancy upward in static fluid ρfV dispg

The table is a recognition aid, not a substitute for a free-body diagram. The direction and magnitude assumptions must still match the physical situation.

11 Worked example 1: angled pull on a rough floor

A 10.0 kg crate is pulled by a rope with tension 50.0 N at 30.0∘ above the horizontal. The kinetic-friction coefficient is μk = 0.20. Find the normal force, kinetic friction, and horizontal acceleration. Use g = 9.81 m/s2.

The vertical force equation is

N +  T sin 𝜃 − mg =  0,
(19)

so

N =  mg −  T sin 𝜃.
(20)

Numerically,

N =  (10.0)(9.81) − (50.0)sin30 ∘ = 73.1 N.
(21)

The kinetic friction is

fk = μkN  = (0.20)(73.1) = 14.6 N.
(22)

Horizontally,

T cos𝜃 − fk = ma,
(23)

so

    50-cos30∘-−-14.6            2
a =       10.0       = 2.87 m ∕s .
(24)

This example shows why N = mg should not be assumed: the upward component of tension reduces the contact force.

12 Worked example 2: spring force and acceleration

A 4.0 kg block is attached to a horizontal spring with spring constant

k = 250 N ∕m.
(25)

The surface is frictionless. The block is displaced 0.080 m to the right of equilibrium. Find the spring force and instantaneous acceleration.

Taking right as positive,

x =  +0.080 m.
(26)

Hooke’s law gives

Fs = − kx = − (250)(0.080) = − 20.0 N.
(27)

Thus

a = Fs-=  −-20.0-=  − 5.0 m ∕s2.
    m      4.0
(28)

The negative signs indicate that both the spring force and acceleration point toward equilibrium.

13 Worked example 3: falling body with quadratic drag

A 2.0 kg body falls downward through still air at 20.0 m/s. Model the drag magnitude as

        2
FD  = cv
(29)

with

c = 0.030  kg∕m.
(30)

Find the instantaneous acceleration and the terminal speed predicted by this model.

The drag acts upward because the velocity relative to the air is downward. Its magnitude is

                  2
FD =  (0.030 )(20.0 ) = 12.0 N.
(31)

The weight magnitude is

mg  = (2.0)(9.81) = 19.62 N.
(32)

Taking downward as positive,

ma  = mg  − FD,
(33)

so

a =  19.62-−-12.0-=  3.81 m ∕s2
         2.0
(34)

downward.

At terminal speed the net force is zero:

mg =  cv2t.
(35)

Therefore

|-----------------------|
|    ∘ mg--             |
vt =   ----=  25.6 m∕s. |
---------c---------------
(36)

14 Practice problems

  1. A 6.0 kg book rests on a horizontal table. Identify the forces on the book and determine the normal force.
  2. A 3.0 kg lamp hangs motionless from one vertical ideal cord. Find the cord tension.
  3. A spring with k = 180 N/m is stretched 0.050 m from equilibrium. Find the spring-force magnitude and state its direction.
  4. A 12 kg crate is pushed horizontally with 30 N and remains at rest. If μs = 0.40, determine the actual static friction and verify that static equilibrium is possible.
  5. A 5.0 kg block slides on a horizontal surface with μk = 0.25. A horizontal force of 25 N pulls it to the right. Find the kinetic friction and acceleration.
  6. A small body moves east at 8.0 m/s through a fluid that itself moves east at 3.0 m/s. If the linear drag law is FD = −bvrel with b = 2.0 kg/s, find the drag force.
  7. A fully submerged object displaces 0.0030 m3 of water of density 1000 kg/m3. Find the buoyant-force magnitude using g = 9.81 m/s2.
  8. A block on a horizontal surface is pulled by a rope angled upward. Explain qualitatively how increasing the upward component of the rope force changes the normal force and therefore the maximum static friction μsN.

15 Answer check

  1. Weight 58.9 N downward and normal force 58.9 N upward.
  2. T = mg = 29.4 N.
  3. Fs = kx = 9.0 N toward equilibrium.
  4. The actual static friction is 30 N opposite the push. Since μsN = (0.40)(12)(9.81) = 47.1 N, static equilibrium is possible.
  5. fk = μkmg = 12.3 N; a = (25 − 12.3)∕5.0 = 2.55 m/s2 to the right.
  6. vrel = 5.0 m/s east, so the drag is 10.0 N west.
  7. FB = ρV g = (1000)(0.0030)(9.81) = 29.4 N upward.
  8. The upward rope component reduces N, so it also reduces the limiting static-friction magnitude μsN.

16 Where the force models go next

This article is intentionally a hub. The following M02 articles develop the force models in more detail:

  • M02-04: weight, normal force, apparent weight, and curved support motion;
  • M02-05: tension and massless strings;
  • M02-06: pulleys and Atwood machines;
  • M02-07: static and kinetic friction;
  • M02-08: inclined-plane dynamics;
  • M02-09: spring force and Hooke’s law;
  • M02-10: drag and terminal velocity;
  • M02-11: dynamics of circular motion.

17 Summary

The most useful force-modeling habit is to identify the interaction before writing a formula. Near Earth’s surface, weight is modeled as mg; a surface can provide normal force and friction; a taut string can provide tension; a spring can provide a restoring force; and a fluid can provide drag and buoyancy. Once the individual forces have been modeled, Newton’s second law combines them:

|------------|
|∑           |
----F-=--ma.--
(37)

References

[1]   PhysicsLibrary, M02-01, Newton’s Laws of Motion.

[2]   PhysicsLibrary, M02-02, Free-Body Diagrams.

[3]   J. Moore et al., Mechanics Map, force and particle-dynamics materials, CC BY-SA 4.0.

[4]   University of California, Davis, Physics 9A Classical Mechanics materials, CC BY-SA 4.0.

[5]   S. J. Ling, J. Sanny, and W. Moebs, University Physics, Volume 1, archived 2016 CC BY 4.0 revision.


"Common Forces in Mechanics" is owned by bloftin.
(view preamble)
View style:
Other names:  M02-03
Keywords:  force, weight, normal force, tension, spring force, Hooke's law, friction, drag, buoyancy, Newtonian mechanics

Attachments:
GRE Physics Companion: Common Forces in Mechanics (Example) by bloftin

Cross-references: terminal velocity, drag force, speed, Free-body diagrams, external force, volume, velocity, dimension, position, equilibrium, displacement, Hooke's law, system, Tension, parameters, vector, static, acceleration, field, mass, The Gravitational Field, drag, friction, formula, magnitude, mechanics, particle, motion, forces

This is version 1 of Common Forces in Mechanics, born on 2026-09-28.
Object id is 1337, canonical name is CommonForcesInMechanics.
Accessed 20 times total.

Classification:
Physics Classification: 45.50.-j (Dynamics and kinematics of a particle and a system of particles)
 45.05.+x (General theory of classical mechanics of discrete systems)
Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:

No messages.

Interact
rate | post | correct | update request | add example | add (any)