File talk:Free body diagram.svg

Genesis of this picture

This drawing is a modified version of File:Free Body Diagram.png. It comes from discussions on AndrewDressel's talk page which are copied below.

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Do you have the source to your diagram Image:Free Body Diagram.png? It needs to be modified, since it shows the frictional force going through its true line of action but the normal reaction going through the block's centroid, which is not, which is somewhat misleading. Thanks, BigBlueFish (talk) 16:08, 9 June 2008 (UTC)

Not handy. I just drew it in MS Word, then pasted it into Paint to save it in PNG format. To where do you propose to move the normal force arrow? Its exact location is not known without additional information and further analysis. -AndrewDressel (talk) 16:26, 9 June 2008 (UTC)

I may stand to be corrected, but since the weight acts through the surface of contact, the normal reaction must act through the point on that surface vertically below the centre of mass, or there would be a net rotational acceleration. BigBlueFish (talk) 16:39, 9 June 2008 (UTC)

That would eliminate any moment due to the gravitational and the normal forces if they were in opposite directions, which they are not, and would not address the moment generated by the friction force. All of this only applies to the static case, which is not given. In fact, if the coefficient of friction is zero, then the normal force is correct as drawn. Otherwise, the acceleration must be known in order to correctly place the single, resultant normal force arrow. -AndrewDressel (talk) 16:50, 9 June 2008 (UTC)

If their lines of action intersect then there is no moment regardless of direction. BigBlueFish (talk) 17:06, 9 June 2008 (UTC)

This contradicts your statement above. Since the normal force is not parallel to the gravitational force, their lines of action will always intersect somewhere, no mater where the normal force is applied. That would mean that the normal force and the gravitational force could never create a moment, which is not true. -AndrewDressel (talk) 19:44, 9 June 2008 (UTC)

The normal and gravitational forces only produce a moment if the weight acts outside the surface of contact. Within this boundary the forces must cross on the surface or the frictional force exerts a moment with the other two. BigBlueFish (talk) 20:57, 9 June 2008 (UTC)

Yes, for a three-force member in static equilibrium. However, there is nothing about the diagram or the article that states that the block is not moving. -AndrewDressel (talk) 01:02, 10 June 2008 (UTC)

Even in a dynamic case the block has no angular acceleration, and all acceleration occurs in a plane perpendicular to the normal reaction, so it is not affected. BigBlueFish (talk) 17:06, 9 June 2008 (UTC)

No angular acceleration is necessary for angular momentum to be a factor. A braking car will experience "weight shift" to the front wheels even if the suspension and the wheels are rigid. No rotation is necessary for this to occur. -AndrewDressel (talk) 19:44, 9 June 2008 (UTC)

Forgive me if I find this a little hand-wavey. I'm not aware of any case in which the vertical force on the wheels of a braking car is imbalanced, or indeed there would be a net moment on the car. A truly rigid car has no angular momentum because its path has an infinite radius of curvature. BigBlueFish (talk) 20:57, 9 June 2008 (UTC)

No hand-waving necessary. It follows directly from the definition of angular momentum about a point, H_p = r_/p x mv, and the correct application of angular momentum balance. You may read about it in the Bicycle and motorcycle dynamics article. The citation includes a link to a PDF copy of a statics and dynamics textbook published by Oxford University Press. -AndrewDressel (talk) 01:02, 10 June 2008 (UTC)

Besides whatever effect or phenomenon you're trying to describe, the simple case that the diagram is trying to describe either shows a block in mechanical equilibrium on a plane or a block sliding down it, both of which are cases in which the free body illustrated experiences no net moment. BigBlueFish (talk) 20:57, 9 June 2008 (UTC)

There is no net moment only in the static (or constant velocity) case. If there is acceleration, then the net moment about any point must equal the cross product of the position vector from that point to the center of mass with the product of mass and acceleration vector of the center of mass:M_p = r_cm/p x ma_cm. Using a point that coincides with the center of mass provides no information, of course, because the position vector has zero length: the net moment about the center of mass is zero.

As for what I am showing in that diagram, it happens to coincide with option c of the three "sensible" ways to represent contact force distributions shown on page 95 of the same textbook mentioned above. One more option I've seen in a different text that is slightly more complicated is to indicate an unknown distance "d" from one edge to the point of application. -AndrewDressel (talk) 01:02, 10 June 2008 (UTC)

I come back to this rather old discussion. I agree with Bigbluefish. If we compute the net momentum at the center of mass, then we find that the object has a rotational acceleration due to ${\vec {F}}_{f}$. If we calculate it at the point of application of ${\vec {N}}$, then we find that the object has a rotational acceleration due to ${\vec {P}}$. Etc. This is not possible because the plane prevents the object from turning. This is far different from the bicycle where the link allows an inclination; a planar link between objects does not allow inclination. (Sorry if the technical vocabulary is not accurate, I'm not a native french speaker.) -cdang|write me 07:51, 4 October 2011 (UTC)

Your analysis is only true if the block is stationary. If the block is accelerating along the ramp in either direction, then the sum of moments about a point will not be zero. I do not understand what you are saying about a bicycle and a link. -AndrewDressel (talk) 21:12, 4 October 2011 (UTC)

I partly agree, what I claim is only true if the frame of reference is inertial, and if the inclined plane is not rotating in this frame of reference; we talk about torque, so this is not a matter of acceleration but of angular acceleration: the m⋅a vector applies at G. Forget about the bicycle, it's not so important. Anyway, nothing in the picture or in the caption tells that the ramp could be rotating, or that the frame of reference is not inertial; thus, your picture is misleading, as most people would consider this. -cdang|write me 08:16, 17 October 2011 (UTC)

I have not assumed an accelerating reference frame or that the ramp is rotating. I have also not assumed that the block is stationary. I should have stated above "If the block is accelerating along the ramp in either direction, then the sum of moments about a point other than the center of mass will not be zero. No rotation, only non-zero acceleration, is necessary for angular momentum of the center of mass about some point that is not the center of mass to be non-zero. Again, I refer you to page 95 of this textbook. I have chosen the simplest of the three options for modeling the unknown distribution of contact forces between a block and a ramp, when the acceleration of the block remains, as yet, unknown. -AndrewDressel (talk) 13:39, 17 October 2011 (UTC)

• The sum of the moments of the forces about any point is always the same: if you change the point, you change the moment arms, but you do not change the sum.
• In point particle mechanics, the Newton's second law of motion states ∑F = ma. In rigid bodies mechanics, you add ∑M_A(F) = Iα, where A is any point, center of mass or not, the result is always the same.
• The sum of the moments of the forces about any point is not zero on your picture, therefore an angular acceleration α.
• The p. 95 of the textbook mentions the existence of a couple; there is no couple drawn on your figure. What we, BigBlueFish and I, suggest, is that you represent the "Equivalent force with no couple" (to refere to the textbook). Or that you represent the couple (turning arrow) that set α to 0. -cdang|write me 15:14, 17 October 2011 (UTC)

• The sum of the moments of the forces about any point is always the rate of change of the angular momentum about that point. For a body translating without rotation, the rate of change of angular momentum about the center of mass is zero, and the rate of change of angular momentum about any other point is only zero if there is no acceleration. Thus the sum of the moments about the center of mass will differ from the sum of the moments about some other point for an accelerating body even in pure translation.
• In rigid body mechanics, Euler's second law of motion states ∑M_A = ^d/_dtH, which in 2D can be expressed as r_cm/A x a_cmm + I α
• That the sum of the moments about a point other than the center of mass is non-zero does not mean that angular acceleration is non-zero, but only that the center of mass is accelerating either translationally or rotationally.
• Page 95 of the textbook states "In 3D ... A couple may be required." Since the block on a ramp can reasonably be modeled in 2D, I see no reason for the existence of a couple.
• You may argue that a different free body diagram example might be more appropriate for pedagogical reasons, but I do not believe that you have yet shown any technical errors with the current version. -AndrewDressel (talk) 17:04, 17 October 2011 (UTC)

Yes I argue that a different free body diagram example is more appropriate for pedagogical reasons, no I did not shown any technical errors with the current version.

Concerning the calculation, the english notation is a bit unsusual for me, and I don't have the time to write some math yet (maybe later). Here is thus a graphical consideration: mind that the N and F_f vectors are just the model of a mechanical action; there is just one mechanical action, the action of the ramp on the object, which can be represented by a single vector R, and this vector can be decomposed in N and F_f, see File:Reaction support avec et sans frottement alt.svg.

Now draw your diagram with just mg and R; if mg and R are not on the same line, then you have a couple, therfore an angular acceleration.

cdang|write me 15:30, 18 October 2011 (UTC)

Andrew's free body diagram represents a classic one that could have come straight out of any number of textbooks. Breaking the force vectors into normal force and frictional force is exactly what I was taught when learning to draw such diagrams. Of course there will be a couple present when N and mg are not parallel, but the fact of its existence does not require that it be shown on this diagram. If you can come up with a better diagram, by all means do so, but understand that this diagram is exactly the form that engineering students in the USA are taught to draw. Ebikeguy (talk) 19:31, 18 October 2011 (UTC)

I'll bite anyway. The current free body diagram shows the general case: the dimensions of the block, the angle of the ramp, the coefficient of friction, and the acceleration of the block are all unspecified. The forces generated by the ramp on the black may be modeled in a variety of ways. If we choose to model them with orthogonal frictional and normal components and then model the normal component as a single resultant force, the actual location where it may be considered to act can only be found in three particular situations without calculation: zero angle, zero friction, or zero acceleration. The current diagram happens to be correct for the middle case, zero friction, but that is not its goal, and it only implies angular acceleration if the translational acceleration is known to be not equal to $g(sin(\theta )-\mu cos(\theta ))$. Instead, this general case shows the use of the free body diagram as an analysis tool, not merely a graphical depiction of some precalculated result. The exact location of a single normal force is not known and so is merely placed in the middle.

Which free body diagram would you wish to show instead, and why? -AndrewDressel (talk) 02:02, 19 October 2011 (UTC)

OK, I uploaded a picture: graphics can show what calculations hide. Watch File:Pfd plan incline avec frottement.svg (please magnify to see the application point of R), esp. the right figure : non-inertial frame of reference link to the object, there is an inertial force F_ie = -ma. In this frame of reference, the object is at the equilibrium which means:

• equilibrium in translation, ∑F = 0, the forces form a triangle;
• equilibrium in rotation, the line of forces all go through the same point G:
  • the weight P applies at the center of gravity G, so its moment about G is 0,
  • the inertial force F_ie applies at the center of inertia which is also G, so its moment about G is 0,
  • the object is not turning so the moment of the action of the plane R about G must also be 0, thus G is on the line of the force R.

This means that R does not apply at the center of the downward face of the object (except for a specific valule of the coefficient of friction μ, which is definetly not the general case):

• in statics, it applies vertically below the centre of mass;
• in dynamics, it applies slightly besides the vertical to the center of mass.

If you decompose R into N and F_f, then you can place F_f anywhere on the plane, but N must be on an action line going through G (and having an angle φ = arctan μ with the perpendicular to the plane, i.e. on the cone of friction). Right?

So I recommend that N applies either at the vertical below G (statics) or slightly besides, but not at the center of the downward face of the object

cdang|write me 09:28, 19 October 2011 (UTC)

I uploaded another picture with a bigger inclination, so things are clearer: File:Pfd plan incline plus avec frottement.svg.

OK, in this case, R is quite far away from the vertical to G, it is closer to the center of the downward face, but it is not right at the center of this face. And this is quite caricatural: I have never seen such a slope in the real life.

cdang|write me 09:54, 19 October 2011 (UTC)

I do not understand what you have drawn at all. I asked "which free body diagram would you wish to show instead, and why?"

• They mistakenly show the ramp, which is supposed to be replaced by the forces it applies to the block in a free body diagram of just the block.
• What equations would you write from these diagrams? R + P + ma = 0? Then what? No coordinate system is provided in the diagram so it is not clear how to decompose any of these vectors.
• How would you find the magnitude of the friction force, the normal force, or the location of where a single normal force may be considered to act from this diagram?
• Do you really propose to introduce free body diagrams with a non-inertial frame of reference?
• The "cone of friction" it includes has not a single hit on Wikipedia, so that will take some further explaining and probably deserves its own article.

Instead, if you are still trying to prove that the normal force may not be considered to act in the middle of the face, why? I do not claim it is. I merely claim that its location cannot be known until further analysis is done, and the middle of the bottom edge is good enough to start. The free body diagram is a tool used to analyze a problem with unknowns, not merely a diagram to depict the results all neat and tidy. -AndrewDressel (talk) 14:57, 19 October 2011 (UTC)

Free-body diagram for a block on a ramp

I did not draw what you should draw (i.e. I did not draw free body diagrams), I just draw a picture that explains what I mean. This type of drawing is usual for low level students (e.g. vocational education) in France, they need to see things and have problems with abstraction (replacing an object by an arrow puzzle them). I don't suggest you to draw the cone (see below on that topic) nor to use a non inertial frame of reference; this non inertial frame of reference makes it easier to understand the problem of rotation (you don't need to introduce the angular momentum), and why R must be where it is.

To be clear, here is what I do not like in your picture:

• the F_f arrow is longer than N, which would mean that μ > 1; if ever this situation exist, it should be rather rare and thus not the general case; please draw F_f smaller than N (half would be quite enough);
• N acts in the middle of the face, which is not the general case.

I don't use equations here, I use graphic statics (statics = the reason why I need to replace ma by F_ie), which is more useful to draw correct diagrams. Here are some basics about graphic statics:

• cone of friction: if F_f = μ×N, then the angle between R = F_f + N (vectors) and the normal to the plane is φ = arctan (F_f/N) = arctan μ; this defines a cone, see File:Adherence traction horizontale.gif;
• R + P - ma = 0 or R + P + F_ie = 0; graphically, the sum of vectors is placing vectors one after each other, 0 means that the distance between the extremities is 0, thus a closed triangle; so if you know P, the direction of a (the ramp) and the direction of R (φ/normal), you can graphically determine the length of R and ma (you know 1 side of the triangle, and the direction of the 2 other sides) without calculation and thus draw a correct diagram even without any value;
• the line of forces must be concurrent in case of 3 forces that are not parallel; see File:Equilibre des forces 3 cables.svg, and search "graphic statics concurrent" (the case of 3 forces is very restrictive, it however allows the solving of numerous problems).

Finally, here is how the diagram should look. You can ask for deletion if you want, I will not use it because I mainly write for French wiki.

cdang|write me 15:54, 19 October 2011 (UTC)

This is all very nice, but it appears that you are confusing the free body diagram, a tool used to set up the equations of equilibrium or motion so that they can be solved for the unknown forces, with some kind of "graphic statics" diagram, a tool apparently used to solve directly for the directions and magnitudes of the unknown forces. I have checked several textbooks, and none mentions using the length of the force arrow to indicate the magnitude of the force, though I am familiar with the concept from graphical methods of vector analysis. Probably in the age of slide rules, such direct graphical techniques were probably better than converting to numbers only to manipulate them by other graphical techniques. In the age of digital calculators and computers, however, I believe the preferred technique is to manipulate the numbers directly, and it is probably far more scalable as well. -AndrewDressel (talk) 01:06, 20 October 2011 (UTC)

(I come back to the left because the indentation starts to be a little bit too big.)

OK, the free body diagram is just used to set up things and not to directly find the solution. However, it should not provide false informations, amongst which:

• F_f cannot be greater than N; the typical coefficient of friction range 0.1 (steel/bronze+lubrication)-0.8 (tyre/dry road), I don't know any μ higher than 1 (although I don't pretend it is impossible);
• the application point of N has no reason to be at the center of the face, this is important to figure out the couple that can be transmitted in a planar joint.

For the latter point, I have another explanation: the action of the ramp on the block is spread all over the surface, and has no reason to be uniform (no symetry). The bottom carries more weight than the top, so the contact pressure is more important at the bottom. N is the integration of this pressure, and is placed at its barycenter; this barycenter is on the down side, not at the center. When a vehicle accelerates, the weight is transfered backwards; this is the same here, so the application point moves towards the top when the block slides, but there is no reason why it should come to the center. It is not impossible, it is just exceptional, and is not likely to occur (unless the slope is very steep).

Placing N at the center is just like drawing an equilateral triangle to explain a general property (like ∑angle = 180°); it is not false, it is just misleading as one could think this is always the case. Got my point?

Concerning graphi statics: yes, it was invented before the calculators and computers. However, it is a powerful tool that allows teaching mechanics to students who have a poor level in math, which is my case. Additionally, it helps thinking, esp. in design: when designing in mechanical engineering, it is more useful to think "moment arm" than "cross product", because the calculations come only once you decided the general architecture.

Yes, calculations are universal and allow a big accuracy, and graphic statics only applies for planar problems and have a limited accuracy. But graphic statics is useful to make illustrations that provide correct information.

cdang|write me 07:20, 20 October 2011 (UTC)

No answer, I considere this discussion as "stable", which does not mean I considere I am right just because I was the last to talk. I copy the discussion on File talk:Free body diagram.svg, so anybody can make his own opinion on the topic.

cdang|write me 14:20, 2 November 2011 (UTC)