The question that the book is asking does depend on the internal stresses, though. The answer for a high-stress state is different from the answer for a low-stress state, and so unless we know what the stress is, we can’t determine the answer.
It literally does not:
The uniform rod AB lies in a vertical plane with its ends resting against the surfaces AC and BC. Knowing that the coefficient of static friction is 0.25 between the rod and each of the surfaces, determine the range of values of θ corresponding to equilibrium when (a) α = 45 °, (b) α = 25°.
The question has nothing to do with stress (either internal or at the contact points), and a rigid element cannot have any internal stress by definition because it cannot be deformed. The member AB isn’t “jammed in”, it doesn’t have any loads at the contact points other than the normal force of the weight of the rod and the static friction force. Continuing to claim that the answer to this problem (and the previous truss problem above) cannot be answered because you don’t accept the assumptions of the problem is just arguing for the sake of argument instead of some real pedagogical purpose. The purpose of this problem is to illustrate how to consider the effects of multiple frictional contacts, and the member and walls are assumed to be rigid because a) it makes it easier to focus on the principal effect of static friction, b) doesn’t introduce a bunch of more advanced concepts that the student has not yet been exposed to, and c) because this is a text for a statics course dealing with rigid bodies.
Yes, all real world structures have some degree of compliance because real materials cannot be perfectly rigid, and in some cases that the amount of compliance does affect the behavior of the system being analyzed. But in many cases, including a real round-ended bar “with its ends resting against the surfaces” made of a reasonably stiff material such as aluminum or steel could be treated as effectively rigid for the purpose of this analysis, namely finding the angles at which static friction would keep the bar from moving. We make assumptions like that in structural and dynamic analysis and machine design all the time because trying to include all minor effects of flexibility, hysteresis, damping, et cetera, would create unwieldy solutions or force a finite element model to have such high fidelity that it would be computationally prohibitive to solve in order to represent effects that have a negligible impact upon the resulting behavior.
Stranger
Since the coefficient of friction is given as 0.25, if the bar was jammed in, it would slip at a less extreme angle, releasing the stress, and allowing the angle to be increased to that of the unstressed bar. Although you could argue that once it slips at the lower angle, the kinetic friction coefficient, which is less than the static friction coefficient, would come into play.
When working such problems there are often a number of unwritten assumptions that need to be made, and no preload is typically one of them. If I wanted a pedantic excuse to avoid having to work would this particular problem, instead of complaining about possible preload I would point out that the radius at the ends of the bar as shown in the illustration is not specified.
Are you sure you know what “stress” means? Even a system as simple as a block of solid material sitting on the floor has internal stress. If you’re working with a model that’s so simple that all elements have no internal stress, you can’t do anything at all with that model.
“Rigid” means that an object has no (or infinitesimal) strain, not that it has no stress.
Yes, I know what “stress” means in an engineering context. I also know that the stress is related to strain in a material by the relationship
\sigma = E\times\epsilon
where E is the modulus of elasticity (often referred to as Young’s modulus). Without any strain (deformation) there is no internal stress or preload regardless of the load or pressure profile that is applied. The claim that “If you’re working with a model that’s so simple that all elements have no internal stress, you can’t do anything at all with that model”, is absurdly wrong, as most mechanism synthesis assumes effectively rigid body behavior unless a member is specifically designed for compliance.
You’ve now gone beyond just misinterpreting the problem and confusing the o.p. to presenting incorrect and counterfactual information. Please stop.
Stranger
FWIW, I flipped through the book and, as expected, there are explicit caveats about internal forces and deformations, statically indeterminate structures, etc., and the limitations of the methods used, so it is not like the very possibility is not acknowledged or that they are lying to the student.
To your main question here…
Yes, the idea is that both ends would contribute a force corresponding to their own impending-motion limit, even if one would already be slipping if it were on its own. A way to think of it is that the “already wants to slip” connection is, in fact, not slipping (even if due to other constraints), so the surfaces in contact there are still able to provide their maximum static frictional force (but no more).
For a different context, consider the diagram below:
upside-down U-shaped
rigid block
________________
| ______ | --->
|____| |____|
============++++++++++++++
surface A surface B
Tilt the whole diagram clockwise at an angle \theta so that the surfaces form an inclined plane. The left foot of the block touches surface A and the right foot touches surface B. We can ask “At what angle \theta will the block begin moving?”
First assume the surfaces are identical. This is just a boring “block on an inclined plane” problem, and both the left and right feet help prevent slipping by the same amount, namely \mu F_N each, where F_N here is the half the weight of the block times \cos \theta (half because of the two feet).
Now, if the surfaces are in fact different (as suggested by the diagram), then each foot contributes what it can, with the left foot resisting motion with force \mu_A F_N and the right foot \mu_B F_N. Both should be tallied when finding the net lateral force at the point of impending motion.
Alternatively, if the weight were distributed unevenly across the two feet (e.g., if one foot were closer to the center of mass than the other), then even if surfaces A and B were identical, the two feet would contribute frictional force up to their different individual limits.
At the microscopic level, any exceeding of the lower of the two limits will lead to adjustments at the “failing” interface and/or macroscopically negligible deformations somewhere in the system such that the frictional force being supplied stays within whatever limit that surface-to-surface interface can tolerate. And, thus, one can apply this approach in many applications of rigid objects in static equilibrium.
At the microscopic level of contact between surfaces (which is the domain of tribology) there are a wide array of different phenomena that contribute to the collective effect of “friction”, including mechanical locking, electrostatic interactions, viscosity (for surfaces that have have fluid between them or which are hyperelastic), et cetera. It is generally difficult to characterize these explicitly or model them in a practical way in real world applications, so in nearly all analysis of applications that don’t involve a fluid lubricant under stress, the effect is just simplified to a empirical or semi-empirical function of normal force and relative velocity. For a static application such as the problem of the o.p., determining the possible friction force as a function of a coefficient times the normal force is sufficient, and frankly is used in the vast majority of real world structural and loads analysis.
Your example of a block impending motion on two different surfaces with different coefficients of friction is a good illustration that the maximum potential friction on both contacts has to be exceeded in order for the block to move, and the only clarification I have is that in the problem presented by the o.p., the points A and B would be moving in different directions relative to the intersection of the planes (A toward and B away, or vice versa), but the principle is the same in that both ends have to be at or beyond the static friction they can develop to be impending motion because they are both constrained to slide along those surfaces.
Stranger
If something relevant is unspecified in the initial problem statement, such as preload in the bar, why not solve it in terms of a parameter of unspecified value instead of saying the problem is unsolvable. The weight and length of the bar are not given in the statement of this problem, and they turn out to not be relevant as long as they are both positive. For this problem, it seems obvious that no preload is intended, but it could be solved in terms of a given value of preload, although in that case it would make sense to also include a parameter for the compressiblity of the bar- a sort of spring constant based on the modulus of elasticity and dimensions of the bar.
I had some time to work on the problem tonight, and I think I made some progress. I used BC and AC as my x and y axes, and summed the forces in each direction. And I summed the moments about the center of the rod. I’ve got three equations. I suspect there are some trigonometric rules I can use, but my last trig class was ~40 years ago. Still working on it.
I’ll have to print this out and see if I can follow your reasoning next time I grapple with the problem.
Near the end of the problem I used the trig identity:
2sin(x)cos(y) = sin(x+y) + sin(x-y)
Using CB as the x axis, and CA as the y axis.
Motion is impending to the right, so frictional forces are the left (negative in x, positive in y).
Length of rod is 2h.
Weight of rod is W.
\begin{equation}
\sum F_{x}: N_{a} - F_{b} - W\sin \alpha = 0\\
N_{a} = F_{b} + W\sin \alpha\\
\end{equation}
\begin{equation}
\sum F_{y}: N_{b} + F_{a} - W\cos \alpha = 0\\
N_{b} = -F_{a} + W\cos \alpha\\
N_{b} = -\mu_{s}N_{a} + W\cos \alpha\\
N_{b} = -\mu_{s}\lgroup F_{b} + W\sin \alpha\rgroup + W\cos \alpha\\
N_{b} = -\mu_{s}F_{b} - \mu_{s}W\sin \alpha + W\cos \alpha\\
N_{b} = -\mu_{s}^{2}N_{b} - \mu_{s}W\sin \alpha + W\cos \alpha\\
N_{b} +\mu_{s}^{2}N_{b}= - \mu_{s}W\sin \alpha + W\cos \alpha\\
N_{b} \lgroup1+\mu_{s}^{2}\rgroup= - \mu_{s}W\sin \alpha + W\cos \alpha\\
N_{b} = \frac{- \mu_{s}W\sin \alpha + W\cos \alpha}{1+\mu_{s}^{2}}
\end{equation}
\begin{equation}
\sum M_{A}: -hW\sin\lgroup\alpha+\theta\rgroup+2h N_{B}\sin \theta-2hF_{B}\cos \theta=0\\
-W\sin\lgroup\alpha+\theta\rgroup+2 N_{B}\sin \theta-2F_{B}\cos \theta=0
\end{equation}
Am I on the right track?
(Also, does anyone know to left-justify and number equations in LaTex?)
Not sure about numbering, but try using \begin{eqnarray} instead of \begin{equation}, and then use &=& instead of = to line up on equals sign.
\begin{eqnarray} \sum F_{y}: N_{b} + F_{a} - W\cos \alpha &=& 0\\ N_{b} &=& -F_{a} + W\cos \alpha\\ N_{b} &=& -\mu_{s}N_{a} + W\cos \alpha\\ N_{b} &=& -\mu_{s}\lgroup F_{b} + W\sin \alpha\rgroup + W\cos \alpha\\ N_{b} &=& -\mu_{s}F_{b} - \mu_{s}W\sin \alpha + W\cos \alpha\\ N_{b} &=& -\mu_{s}^{2}N_{b} - \mu_{s}W\sin \alpha + W\cos \alpha\\ N_{b} +\mu_{s}^{2}N_{b}&=& - \mu_{s}W\sin \alpha + W\cos \alpha\\ N_{b} \lgroup1+\mu_{s}^{2}\rgroup&=& - \mu_{s}W\sin \alpha + W\cos \alpha\\ N_{b} &=& \frac{- \mu_{s}W\sin \alpha + W\cos \alpha}{1+\mu_{s}^{2}} \end{eqnarray}
That’s handy; thank you.
I don’t usually solve these sorts of problems by taking the general case and just plugging in values at the end. I know the ocefficient of static friction, and alpha has one of two possible values. I find it a lot easier to work with actual numbers, but since I had @Manlob’s roadmap to follow, I thought I’d try the general case this time. And it’s been kinda interesting to do all those equations in LaTex, and also frustrating. I found a nice little editor here: LaTex Equation Editor | Tutorialspoint
Oh, I was a bit sloppy with my subscripts. I should have used capital A and B for the normal and frictional forces at those points. If that’s the only mistake I made, I’ll be pretty happy.
Yes, that looks good. Another tip is go back to the trig from a couple years ago near the beginning of this thread:
mu/sqrt(1+mu^2) = sin(atan(mu)) and 1/sqrt(1+mu^2) = cos(atan(mu))
I don’t quite see how that helps me. Looks like I’m just going to have square roots all over the place.
When you put your expression for Nb into the moment equation it will contain something like:
(sin - mu cos)(-mu sin + cos)/(1+mu^2)
The 1+mu^2 in denominator is two of the square root terms multiplied together.
The solution manual is kind of confusing, but there is a nice solution implied there. You can boil the problem down to a rectangle of vectors. Sorry it isn’t easy to post an image here. Two sides of the rectangle are the resultant force vector at Point A, and the other sides are the resultant force vector acting at Point B. One diagonal is a vector, W, pointing straight down and is the weight of the bar, with a triangle of vectors representing force equilibrium: 0=W+Ra+Rb. Moment equilibrium is ensured because W, Ra, and Rb, are all directed through the same point. The other diagonal is parallel to the bar. Once you have that rectangle, it is pretty easy to find the angles the diagonals make with the sides and with each other to determine theta in terms of alpha and atan(mu).
\begin{equation}
N_{b} = \frac{- \mu_{s}W\sin \alpha + W\cos \alpha}{1+\mu_{s}^{2}}
\end{equation}
\begin{equation}
-W\sin\lgroup\alpha+\theta\rgroup+2 N_{B}\sin \theta-2F_{B}\cos \theta=0\\
-W\sin\lgroup\alpha+\theta\rgroup+2 N_{B}\sin \theta-2\mu_{s}N_{B}\cos \theta=0\\
-W\sin\lgroup\alpha+\theta\rgroup+2\sin \theta\lgroup\frac{- \mu_{s}W\sin \alpha + W\cos \alpha}{1+\mu_{s}^{2}}\rgroup-2\mu_{s}\cos \theta\lgroup\frac{- \mu_{s}W\sin \alpha + W\cos \alpha}{1+\mu_{s}^{2}}\rgroup=0\\
-W\sin\lgroup\alpha+\theta\rgroup+\lgroup\frac{-2\mu_{s}W\sin \theta \sin \alpha + 2W\sin \theta\cos \alpha}{1+\mu_{s}^{2}}\rgroup-\lgroup\frac{-2 \mu_{s}^{2}W\cos \theta\sin \alpha + 2\mu_{s}W\cos \theta\cos \alpha}{1+\mu_{s}^{2}}\rgroup=0\\
-W\sin\lgroup\alpha+\theta\rgroup-\frac{2\mu_{s}W\sin \theta \sin{\alpha}}{1+\mu_{s}^{2}}+\frac{2W\sin \theta \cos{\alpha}}{1+\mu_{s}^{2}}+\frac{2\mu_{s}^{2}W\cos \theta \sin{\alpha}}{1+\mu_{s}^{2}}-\frac{2\mu_{s}W\cos \theta \cos{\alpha}}{1+\mu_{s}^{2}}=0\\
\end{equation}
I’m sure there’s a point at which these equations start simplifying, but I’m just not seeing it. The closest I ever had to a pure trigonometry class was in 11th grade, and I don’t remember it covering any of these sorts of identities or equivalencies.
Define phi by:
\begin{equation}\phi=\tan^{-1}(\mu_s)\end{equation}
Your moment equation is
\begin{equation}0=-W\sin(\alpha+\theta)+2N_B(\sin\theta-\mu_s\cos\theta)\end{equation}
Using your expression for N_B, it becomes
\begin{eqnarray}
0&=&-W\sin(\alpha+\theta)+2W(-\mu_{s}\sin\alpha+\cos\alpha)(\sin\theta-\mu_s\cos\theta)/(1+\mu_s^2)\\
0&=&-\sin(\alpha+\theta)+2(-\sin\phi\sin\alpha+\cos\phi\cos\alpha)(\cos\phi\sin\theta-\sin\phi\cos\theta)
\end{eqnarray}
From there you can make use of the identies for cos(x+y) and sin(x-y) to simplify. Often when a problem gets very messy and boils down to a simple answer, there may an easier way to do it.
Just wanted to weigh in on Stranger vs Chronos. I’m a licensed civil engineer. My BS and MS are from one of the top civil schools in the USA. Stranger is absolutely correct about how this is taught in an intro to statics level and the rigid assumption is damn useful in many many situations. A civil engineer generally would have a full semester course on that sort of thing (statics). Then another full semester course (mechanics) taking into account internal forces, deformations, compliance, etc.
As a civil engineer, further courses past that point might be structural system design, or specific building materials (steel, concrete, wood) and associated design specifics / building code compliance. Mechanical engineers would probably diverge after mechanics and dynamics courses, but the full amount of real world complexity absolutely isn’t needed in an intro course. My cites are the Willis Tower and Burj Khalifa.