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h1. Force
{excerpt}Historically, the first mathematical description of [interactions|interaction] in Newtonian Mechanics. Forces produce a change in the motion of a mass on which they act, according to F=ma ([Newton's Second Law]), which is a vector equation (the acceleration is in the direction of the net force). Forces result from several types of physical interactions, which always exert _opposite_ forces on two _different objects_ ([Newton's Third Law]).{excerpt}
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h3. {toggle-cloak:id=mot}Motivation for Concept
{cloak:id=mot}
Consider a bowling ball (or some other heavy object that moves with little resistance). If you want the ball to move, you have to exert a force on it in the direction you want it to move. If you want the moving ball to turn, you have to exert a force on it toward the side you want it to turn toward. If you want the ball to stop moving, you have to exert a force opposite to its velocity. To change the motion of the bowling ball, you will probably apply a force by using your hands or feet or some object you push against the ball. There are other kinds of forces, however. The earth, for example, can alter the ball's motion through the invisible action-at-a-distance of [gravity|gravity], often represented as a gravitational field acting on the body at the site of the body.
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h3. Newton's Laws
Newton's famous Three Laws of Motion together comprise his definition of force.
h6. [Newton's First Law]: {excerpt-include:Newton's First Law}
h6. [Newton's Second Law]: {excerpt-include:Newton's Second Law}
h6. [Newton's Third Law]: {excerpt-include:Newton's Third Law}
h3. Classification of Forces
There are many ways to classify forces. For the purposes of the modeling approach to physics, the most important classifications to understand are Internal vs. External and Conservative vs. Non-Conservative. Another commonly encountered classification of forces is by their status as "fundamental" vs. phenomenological.
h4. Internal vs. External
h6. [Internal Forces|internal force]: {excerpt-include:internal force}
h6. [External Forces|external force]: {excerpt-include:external force}
h4. Conservative vs. Non-Conservative
h6. [Conservative Forces|conservative force]: {excerpt-include:conservative force}
h6. [Non-Conservative Forces|non-conservative force]: {excerpt-include:non-conservative force}
h4. Fundamental vs. Phenomenological
h6. [Fundamental Forces|fundamental force]: {excerpt-include:fundamental force}
h6. [Phenomenological Forces|phenomenological force]: {excerpt-include:phenomenological force}
h3. {toggle-cloak:id=app} Application of Newton's Laws
{cloak:id=app}
h4. {toggle-cloak:id=fd} Force Diagram
{cloak:id=fd}
Before setting up the equations of Newton's Second Law for an object, it is vital to quantitatively understand the forces acting on the object. The first step is to draw a [force diagram|force diagram] for the situation, which is a sketch of the situation showing all relevant forces and their point of application. Here is an example:
!forcephysrep1.png!
This example shows a box that is acted upon by each of the five common categories of forces in introductory physics. The person pushing the box applies a contact force _F{_}{~}p~. (The subscript "p" is chosen here to denote "push". You might also use "a" for applied, "c" for contact, or any other subscript that has meaning for you.) The person in front pulls on a rope, which transmits a tension force _T_ to the box. The earth's gravitational force gives the box a weight force _W_. The floor provides both a normal force _N_ and a friction force _F{_}{~}f~. Note that the physical representation should always include a [coordinate system|coordinate system].
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h4. {toggle-cloak:id=fbd} Free Body Diagram
{cloak:id=fbd}
This physical representation is not very useful as a guide to using the equations of Newton's Second Law. If we wish to write Newton's 2nd Law algebraically, it is important to find the components of each vector. An alternate graphical representation that leads naturally to finding the vector components is the [free body diagram|free body diagram]. In a free body diagram, the center of mass of the box is represented as a point at the origin of a coordinate system. All the forces acting on the box are then drawn as [vectors|Introduction to Vectors] with their [tail] at the origin. For the example of the box, the free body diagram would be:
!forcefbd1.png!
which leads naturally to the resolution into components of any vectors that do not point along an axis:
!forcecomp1.png!
{warning}Beware of a temptation to say that in the free body diagram for this example that _N_ is the y-component of _T_ and _F{_}{~}p~ is the x-component of _T_. This is one drawback to the traditional way of drawing free body diagrams. The physical representation above and the vector addition diagrams below should make it clear that these forces are essentially independent (though they are related by Newton's 2nd Law).
{warning}
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h4. {toggle-cloak:id=seclaw} applying Newton's 2nd Law
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Once the free body diagram is complete, Newton's 2nd Law can be applied. It is important to note that due to its vector nature, Newton's 2nd Law is really three laws: one for each coordinate direction. Often, as is the case in our example, there are no forces in the _z_ direction (out of the plane of motion - often the computer monitor).
It is important to note that we have a choice of which directions to take as the x and y axes (although they must be perpendicular). It is wise to exploit this choice to make the resulting equations as simple as possible. Generally it is best to pick the coordinates so that unknown quantities (and particularly an unknown acceleration) lie along just one coordinate axis. This will mean that the resulting equations will give expressions for the unknowns that don't have cos and sin coefficients in front of them.
{latex}\begin{large} \[ F_{p,x} + T_{x} + W_{x}+N_{x} + F_{f,x} = ma_{x} \]
\[ F_{p,y} + T_{y} + W_{y}+N_{y} + F_{f,y} = ma_{y}\]\end{large}{latex}
At this stage, we have made *no use* of the free body diagram. We have just blindly listed all five forces and added them up. The free body diagram is used to simplify these equations. For example, we can see from the diagram that many of the components are zero. Using this information, the equations simplify to:
{latex}\begin{large}\[ F_{p,x} + T_{x} + F_{f,x} = ma_{x}\]
\[ T_{y} + W_{y} + N_{y} = ma_{y} \]\end{large}{latex}
There are two important things to note here. First, only the tension _T_ appears in both equations, because it is the only vector in our diagram that is not directed along a coordinate axis. Second, all the force components are listed with a plus sign, even though our diagram makes it clear that _F{_}{~}f,x~ and _W{_}{~}y~ will be negative. This is not a mistake. Solving the equations would give negative values for these components. Explicitly including the negatives in the equation would mean that the algebra would actually give positive values for these two components.
Since some students prefer to explicitly assign the negatives in the equation, there is another way to write these equations. They can also be written in terms of the [magnitude|magnitude] of each force. An equivalent set for our example is:
{latex}\begin{large}\[ F_{p} + T\cos\theta - F_{f} = ma_{x} \]
\[ T\sin\theta - W + N = ma_{y}\]\end{large}{latex}
Because the negatives have been explicitly included, all quantities will emerge as positive numbers, which is appropriate for magnitudes (but *not appropriate* for vector components\!).
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h4. {toggle-cloak:id=const} Applying Constraints
{cloak:id=const}
Successful application of Newton's 2nd Law often requires an understanding of constraints on the acceleration of an object. For instance, in the example above, it is likely that you would be told to assume the box slides along the floor. In that case, it is clear that the box will not move in the _y_ direction at all. If the box does not move in the _y_ direction, then it certainly has no y-acceleration. Thus, the y-equation should be modified to:
{latex}\begin{large}\[ T_{y} + W_{y} + N_{y} = 0 \]\end{large}{latex}
or, equivalently:
{latex}\begin{large} $ T\sin\theta - W + N = 0 $\end{large}{latex}
{note}It is important to note that acceleration can be zero even if an object _is_ moving. If you were told in the example above that the box was sliding with *constant velocity* along the floor, then *both* _a{_}{~}x~ *and* _a{_}{~}y~ will be zero.
{note}
{cloak:const}
h4. {toggle-cloak:id=net} Graphical Representation of Net Force
{cloak:id=net}
The [free body diagram|free body diagram] is useful for finding vector components, but it can sometimes confuse the concept of net force. The net (or total) force acting on a body is the vector sum of all the individual forces, and this quantity is equal to the mass times the acceleration. Graphically, a vector sum is accomplished by arranging the vectors [tail-to-tip], but a free body diagram arranges all the vectors with their [tail] at the origin. Thus, the free body diagram does not clearly show the net force. It is possible to draw a graphical representation of the net force by arranging the vectors tail-to-tip, beginning at the origin.
!forcenetpic1a.png!
Note that the order of the vectors is irrelevant (the answer will be the same no matter how they are arranged).
!forcenetpic1b.png!
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h3. {toggle-cloak:id=sys} Newton's Laws Applied to Systems
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h4. {toggle-cloak:id=int} Internal Forces and Newton's 3rd Law
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{anchor:internal}
Consider a situation like that shown here, where there is negligible friction from the ground, but the boxes produce friction on each other:
!forcesystem2.png!
We can choose to view the two blocks independently or as a single [system|system]. Suppose that we first choose to consider the blocks as independent objects. In that case, a separate free body diagram must be made for each block. Shown here are [force diagrams|force diagram] and [free body diagrams|free body diagram] for each object:
|| Force Diagram, Box A || Free Body Diagram, Box A ||
| !forcephyspic2a.png! | !forcefbd2a.png! |
|| Force Diagram, Box B || Free Body Diagram, Box B ||
| !forcephyspic2b.png! | !forcefbd2b.png! |
{note}Note in box A's force diagram we have removed objects or people that do not interact with box A. The ground pushes up on box B, which in turn pushes up on box A, but the ground does not _directly_ push on box A. The person pushes on box B, which in turn attempts to move box A, but the person does not _directly_ push on box A.
{note}
We have labeled two of the forces in a new way. The [friction|friction] force and the [normal force|normal force] that act between the objects have been labeled with two subscripts each. Each of the subscripts include the labels of the blocks. The order of the subscripts conveys meaning:
_F{_}{~}AB~ = "the (friction) force *on* A *from* B"
_N{_}{~}BA~ = "the normal force *on* B *from* A"
{note}Note that when double subscripts are used *in this wiki*, the best way to remember their meaning is to insert the word "from" between them.
{note}
{warning}Most physics texts use our convention for subscripts, although a minority use the opposite convention. So *beware*: some of your texts or other reference materials you may use the "wrong" convention.
{warning}
Our inclusion of these subscripts has a point. When you are constructing free body diagrams for more than one object, [Newton's 3rd Law|Newton's Third Law] will apply to any forces that are exerted on one of those objects by another of the objects. Thus, in our example, any forces from block A on block B or from block B on block A will satisfy Newton's 3rd Law. When the double subscript notation is used, Newton's 3rd Law is easy to write:
{latex}\begin{large}\vec{F}_{AB} = -\vec{F}_{BA}\]\end{large}
{latex}
which is usually applied component by component,
{latex}
\begin
{large}\[ F_{AB,x} = -F_{BA,x} \]\[F_{AB,y} = - F_{BA,y}\]\[ N_{AB,x} = - N_{BA,x}\]\[N_{AB,y} = -N_{BA,y}\]\end{large}
{latex}
In other words, for the *force* or its *vector components*, when the order of the subscripts is reversed, a negative sign applies.
{note}Note that we wrote the _y_ direction equation for the friction force and the _x_ direction equation for the normal force even though these components are clearly zero. The reason is to reinforce that Newton's 3rd Law constrains both the relative magnitude _and_ the relative _direction_ of the forces. In terms of magnitudes only, the relationship is: _F{_}{~}AB~ = _F{_}{~}BA~ (the magnitudes are the same, even though the directions are different)\!
{note}
For our example, we can write the equations of Newton's 2nd Law for each object as (removing components that are zero for clarity):
{latex}
\begin
{large}\[ F_{AB,x} = m_{A}a_{A,x}\]
\[ F_{x} + F_{BA,x} = m_{B}a_{B,x}\]
\[ W_{A,y} + N_{AB,y} = m_{A}a_{A,y} = 0\]
\[ W_{B,y} + N_{y} +N_{BA,y} = m_{B}a_{B,y} = 0\]\end{large}
{latex}
{note}For the double subscript forces, only forces with the first subscript the same as the object whose mass is on the right hand side should be included\!
{note}
Substituting from Newton's 3rd Law lets us alter the equations for object A:
{latex}
\begin
{large}\[ -F_{BA,x} = m_{A}a_{A,x}\]
\[W_{A,y} - N_{BA,y} = 0 \]\end{large}
{latex}
{note}Two notes. First, we could just as simply have substituted into B's equations, but A's are shorter. (Substituting into *both* A's and B's equations is foolish - can you see why?) Second, *after* the substitution, all the two-subscript forces in the equations for A are "backwards" (B is the first subscript).
{note}
After the substitution, adding the equations for the _x_ direction and those for the _y_ direction gives:
{latex}
\begin
{large}\[ F_{x} = m_{A}a_{A,x}+m_{B}a_{B,x} \]
\[ W_{A,y} + W_{B,y} + N_{y} = 0\]\end{large}
{latex}
We have constructed two equations from which all two-subscript forces have dropped out\!
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h4. {toggle-cloak:id=ext} External Forces and the System Viewpoint
{cloak:id=ext}
{anchor:external}
It is no accident that the two-subscript forces are so easy to eliminate from the equations of Newton's 2nd Law. It is a simple consequence of Newton's 3rd Law. The fact that the presence of _F{_}{~}BA,x~ in object B's equations guarantees the presence of _F{_}{~}AB,x~ in object A's, and that _F{_}{~}BA,x~ = - _F{_}{~}AB,x~ makes it automatic that these forces will cancel. For this reason, it is possible to skip some of the automatic algebra by taking the system viewpoint.
In the system viewpoint, if we are told that box A does not slip along box B, we might as well treat them as _one object_. In that case, the forces from A on B and vice versa are applied _internal_ to that object. In other words, any forces that would have to be "two-subscript" forces can be described as [internal forces|internal force]. These forces will _cancel_ from the equations of motion of the complete system as a consequence of Newton's 3rd Law. Thus, when writing Newton's 2nd Law for the composite A+B object, we can completely ignore the internal forces. The free body diagram for the composite object, then, is constructed as shown here:
!forcefbdab21.png!
The forces that remain in the free body diagram after internal forces are removed are called [external forces|external force]. As the name implies, each of these forces must involve an object outside the system. In our example, the force from the person on block B clearly involves the person, who is not part of the two block system. The normal force from the ground on block B involves the ground. The weight of each block involves the earth.
We can now write the equations of Newton's 2nd Law:
{latex}
\begin
{large}\[ F_{x} = m_{A+B}a_{x}\]
\[ W_{A+B,y} + N_{y} = m_{A+B}a_{y} = 0\]\end{large}
{latex}
If we use the fact that the [mass] and the [weight] of a composite object is simply equal to the sum of the mass and weight (respectively) of the parts, we have recovered exactly the same equations as we did by considering box A and box B as separate objects, if we remember that we have assumed A and B each have the same acceleration (_a{_}{~}A,x~ = _a{_}{~}B,x~).
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h4. {toggle-cloak:id=cm} Center of Mass
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The system viewpoint can actually be applied even to a collection of objects that are not fixed relative to each other. Suppose we make a minor change to the example above by assuming that block A begins to slide back along the top of block B when the force _F_ is applied. Now, the x-equation from the individual free body diagrams:
{latex}
\begin
{large}\[ F_{x} = m_{A}a_{A,x} + m_{B}a_{B,x} \]\end{large}
{latex}
is in conflict with the system version:
{latex}
\begin
{large}\[ F_{x} = m_{A+B} a_{x}\]\end{large}
{latex}
Clearly, the system version is in error, since the objects should not have the same acceleration\! It can be fixed, however, if we state that the system acceleration should be defined as the acceleration of the system's [center of mass b]. The position of the center of mass is defined as:
{latex}
\begin
{large}\[ x_{cm} \equiv \frac{m_{A}x_{A} + m_{B}x_{B}}{m_{A}+m_{B}} \] \end{large}
{latex}
Taking two time derivatives, then (assuming the masses are constant) immediately gives:
{latex}
\begin
{large}\[ a_{cm} = \frac{m_{A}a_{A,x} + m_{B}a_{B,x}}{m_{A}+m_{B}} \] \end{large}
{latex}
Thus, if we replace the acceleration in the system equation by _a{_}{~}cm~, we find:
{latex}
\begin
{large}\[ F_{x} = m_{A+B} a_{cm} = m_{A+B}\frac{m_{A}a_{A,x} + m_{B}a_{B,x}}{m_{A}+m_{B}} =
m_{A}a_{A,x}+m_{B}a_{B,x} \]\end{large}
{latex}
in complete agreement with the result of using individual free body diagrams. This realization leads us to a general statement of Newton's 2nd Law applied to a system with constant mass.
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h4. {toggle-cloak:id=secsys} Statement of the Law for a System
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For a system with constant mass, we have:
{latex}
\begin
{large}\[ \sum_{k=1}^{N_{F}} \vec{F}^{ext}_{k} = m_{\rm system}\vec{a}_{cm}\]\end{large}
{latex}
It is important to note that the sum is only over _external_ forces. Internal forces will _always_ cancel as a result of Newton's 3rd Law. You can show using similar arguments that the most general form of Newton's 2nd Law is:
{latex}
\begin
{large}\[ \sum_{k=1}^{N_{F}} \vec{F}^{ext}_{k} = \frac{d\vec{p}_{cm}}{dt} \]\end{large}
{latex}
where
{latex}
\begin
{large}\[ \vec{p}_{cm} \equiv m_{\rm system}\vec{v}_{cm} \]\end{large}
{latex}\\
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h3. {toggle-cloak:id=ex} Relevant Examples
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{contentbylabel:example_problem,dynamics|operator=AND|showSpace=false|excerpt=true|maxResults=50}
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