Can friction or normal force be undetermined?

Question

I've been studying newton-mechanics and encountered this weird situation where friction and normal force can't be decided.

It's a problem where a uniform rod is leaning on the wall at degree $\theta$. Rod's length is L, mass is M.

Assume that the ground and wall's coefficient of friction is sufficiently large.

There are two points where the rod takes force, the point where the rod meets the ground, and the point where the rod meets the wall. Let's call these points point P, point Q respectively.

The problem is that if the ground and wall both has friction, the friction and normal force is undetermined.

There are three equilibrium equations available.

equilibrium regarding x axis : $f_1-N_2=0$

equilibrium regarding y axis : $N_1-Mg+f_2=0$

torque equilibrium about point P : $\frac{L}{2}Mg\cos(\theta)-LN_2\sin(\theta)-Lf_2\cos(\theta)=0$

There are four unknown variables $N_1, N_2, f_1, f_2$ but only three equilibrium equations, thus friction and normal force is undetermined.(torque equilibrium about point Q is meaningless, as it can be derived from three equations above)

As far as i know, this result doesn't make sence. Simply thinking, we could put a few pressure sensors at point P, Q to find $N_1, N_2, f_1, f_2$.

What's the real answer? can we derive more equations and find $N_1, N_2, f_1, f_2$? Or is $N_1, N_2, f_1, f_2$ really undetermined?

Answer 1

In general, there is no reason why all the reaction forces can be determined in a problem. Your intuition is biased by all the exercises you were assigned in an academic setting, which were hand-picked to be fully solvable. In this example, intuitively, you have three equations for four unknowns so modulo some mathematical caveats, you cannot hope to find a unique solution.

Essentially, what this means is that your model is incomplete, you'll need additional information. Typically, in a realistic setting, you would say that $f_1,f_2$ arise from Coulomb friction, so the problem would still be under determined and multiple equilibrium positions exist (you can easily check it in real life). The simplest, fully solvable setting would be to assume that $f_1,f_2$ result from two springs. For notations, let's take the coordinate system to be directed by the floor ($x$) and wall ($y$), the rigidity of the spring at $P$ as $k_1$ linking it to the equilibrium point at $(x_0,0)$ and for the spring at $Q$ (the opposing extremity) as $k_2$ linking it to the equilibrium point $(0,y_0)$.

The problem is now easily solvable. The fastest route is by an energy method: $$ E = \frac{k_1}{2}(L\cos\theta-x_0)^2+\frac{k_2}{2}(L\sin\theta-y_0)^2+mg\frac{L}{2}\sin\theta $$ from which you can calculate $\theta$ by minimizing $E$ and deduce the four forces using: $$ f_1 = k_1(x_0-L\cos\theta)\\ f_2 = k_2(y_0-L\sin\theta) \\ N_1 = -mg-f_2 \\ N_2 = -f_1 $$

Notice that the final forces will depend crucially on $k_1,k_2,x_0,y_0$, which means that these extra parameters were needed to fully solve the problem. In particular, you should retrieve the Coulomb friction when $k_1,k_2\to \infty$ and $x_0,y_0$ the initial position of the beam, you'll notice that the indeterminacy is caused by the dependence on the ratio $k_1/k_2$ which is still not determined in this limit.

You can reconcile the previous argument with your intuition regarding a theoretical pressure sensor. It indicates that the reading of the pressure sensor will depend on the physical characteristics of the sensor, not only on the situation. In general, the fact that you can "cancel out" the influence of the measurement device is not trivial and results from a carefully chosen experimental setup.

There exists some simpler examples where you have this kind of indeterminacy. Take for example a rigid horizontal beam that rests on three rigid pillars of the same height, check out Arnold's Mathematical Methods of Classical Mechanics for more.

Hope this helps and tell me if you need more details.