Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We're finding forces (and whether they're in tension or compression) using the Method of Joints and Method of Sections. I don't understand why sometimes it's necessary to find the support reactions, but sometimes it's not. Please help. Thank you!

share|cite|improve this question
My guess is that nobody's answered yet (after 9 hours) because we don't understand the engineering jargon. I don't, at least. If you can find a way to translate to more physicsy language we might be able to help more. – Mark Eichenlaub Mar 19 '11 at 7:02
I agree with Mark. A brief description of a specific example would probably do the trick. – Ted Bunn Mar 19 '11 at 14:44
Sincerely, and with all due respect to the people here that know far more than I do, I must say that if someone doesn't understand the question, I probably would rather wait for someone who is familiar with the terminology. I'm sure you can understand. I accept the possibility that this question may be outside the scope of this site, and I appreciate the comments. – ChrisC70 Mar 19 '11 at 17:27

I have examined some material on Statics and having recently discovered this unanswered question I can provide some indications from a physics perspective of what is going on.

The fundamental issue in statics is to provide the values for the forces acting on a rigid construction (usually of rods) with everything in equilibrium. (The examples I have looked at are 2 dimensional, so I will assume that in what follows; but one could add similar equations for 3 D although the essence of the OP question can be discussed in 2D.)

The equations available are from the Newtonian conservation laws:

$\Sigma F_x = 0$ - the sum of forces in the x-axis is zero

$\Sigma F_y = 0$ - the sum of forces in the y-axis is zero

$\Sigma M_a = 0$ - the moment (torque) about a point (hence any point) is zero

So these equations will be true at any point, especially connection points on the construction (often called a "truss").

The variables that form in these equations are the Forces at given points on the system: more accurately the x- and y- components of the Forces.

The system (truss) will be placed on a ground at N points; at each of these points we have a "support reaction" onto the truss. Thus there will be N unknown support reactions (in the y-axis say) to find in principle: labeled $V_1$, $V_2$,....,$V_N$ say.

There will also likely to be some form of external force in the problem. This external Force $F$ is normally assumed applied at a specific point on the structure, and may have a horizontal (x-axis) component and a downward vertical (-ve y-axis component). I should remark here that in a wider class of physics and engineering problems the force F is assumed to be distributed across the structure and not located just at a point: the primary example of such a force would be gravity. However the mathematical techniques to solve this would involve calculus and so are outside of the basic Statics theory I have seen and that is covered here. So assume a single force is acting at a point externally. Can this problem be solved for all the forces?

Well it turns out that there is a mathematical problem here, which I shall explain with some simple examples.

Example 1: Beam with 2 supports and external force

Let the length of the beam be L units, supports (A,B) at each end and an external force F acting purely vertically on the beam at a point distance $a$ from A, $b$ from B (hence $L=a+b$). Then we have to solve (signs are important in general):

[1] $\Sigma F_y$ = $V_A + V_B - F = 0$

[2] $\Sigma M_A$ = $aF - (a + b) V_B = 0$

Here $F$, $a$ and $b=L-a$ are known and $V_A$ and $V_B$ are unknown. The key point here is that we have two equations with two unknowns. The solution is easily seen to be:

$V_A = (b/a+b) F = (b/L)F$

$V_B = (a/a+b) F = (a/L)F$

So this is called a statically determinate system. However the simplest of modifications can result in a statically indeterminate system.

Example 2: Beam with 3 supports

I wont go through the details which are in this Wikipedia article. The point is that there are the same equations except that there is now a third reaction force $V_C$. Therefore there are too many variables and this problem cannot be solved (by Statics alone). In a sense this is quite a general case.

However the engineering challenge remains to model the forces on a connected truss. The Method of Joints does this going through the Truss iteratively: at each connection point with just a few forces and variables such that the problem is determinate and such that a new unknown force is determined. Eventually by repeated application of this technique all the forces on all the beams of the truss can be determined.

A different question can always be asked about any physical system however: namely what are the forces acting on a particular point? This problem does not always need the full solution of the forces acting on every other point determined before it can be answered. So the idea is to construct a Free Body Diagram which extracts the key physics (in this case Statics) of the problem. Doing so in Static Truss like problems is called the Method of Sections. The key equations are the Newton equations of Force and Torque used in Statics, but applied in a selective way.

This technique of determining where to "cut" the truss looks like a bit of an art, although there do seem to be some principles used in the Tutorial material I have seen. Here is one such Trusses - Method of Sections .

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.