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Hello, I am Civil engineer Bilgehan Yılmaz.

First of all, I would like to ask a question about basic statics, I modelled my question below. The following forces are applied around a wooden stick with a weight of $100N$. For an object to be in equilibrium, the total force and total moment must be zero.

In Prob-1, I have shown a frictional force that will prevent the rod from rotating around $D$ point, but in this case, the total net force on the object will move to the left as $300N$ and the moment will be different from zero everywhere on the object except at the $D$ point.

In prob-2, the net force on the object is zero. However, the object rotates around $D$ point.

  • When writing the equilibrium equations of the object, is the friction force taken to make the total moment at the center of the object zero $(Prob-1)$ or is it taken to make the total net force zero $(Prob-2)$? that is, does a frictional force always consist of the net force on the object? Or is there no friction caused by the moment on the object to prevent it from turning?
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  • $\begingroup$ It will not make a difference to the final answer but your diagrams are missing a force: the force due to gravity. $\endgroup$
    – JohnHoltz
    Jun 22, 2023 at 23:43
  • $\begingroup$ Yes, I didn't draw it on purpose. $\endgroup$ Jun 23, 2023 at 6:02

1 Answer 1

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Premise

When you analyze the equilibrium of a body you first have to be sure that it is in a static configuration. In other words, it's bound to enough external constraints such that the reaction forces can always (ideally) counteract the action of any external force or momentum applied to the body.
An example on the picture below:

As you can see, the beam on top is said to be in a static configuration because it can counteract the action of any force or momentum by combining the reaction forces of the two constraints: a horizontal and a vertical reaction force on the left constraint, and a vertical force on the right constraint. The beam on the bottom is not static by definition. It could be in a static situation if the external forces balance out in the horizontal axis, but most probably it will be in a dynamic situation.
Now your situation is more similar to something like this:

And since the body is not glued to the ground, you have the situation on the center, so the body can't translate to the left-right due to friction, or up-down due to the ground, but it can still rotate. Being this the situation, you cannot expect to have static equilibrium each time. Sometimes you will be in a dynamic situation, as friction force can only constrain the body so much. It's also possible that the forces are so strong that the friction force is overcome, and then the body will be as in the situation on the right, where only the ground can help in constraining it by a vertical force only.

Equilibrium of the wooden stick

Here we are in a dynamic case, so the friction force won't be calculated using the static equilibrium formulas. You need to take into account the accelerations of the system and the dynamic forces produced by it.

But before rushing to the formulas, let's start by considering what could happen. The following cases are possible outcomes:

  • Friction force is strong enough to keep the right corner still. The block will pivot around the corner;
  • Friction force is not strong enough. The block will slide and roll to the left;

We need to keep these in mind, as the procedure to find the two forces will vary.

Kinematic description

Let's start by defining two reference systems like in the picture above. The black one is an inertial reference system, the red one is fixed to the body and rotates with it.

The formulas that describe the motion of point D and G and rotation about the center of mass of the stick at the initial instant are:

$$ \left\{ \begin{array}{c} \ddot{x}_D = \ddot{x}_G - \omega^2 x'_D - \dot{\omega} y'_D = \frac{ \sum_{i} F_{x,i}^{ext} }{m} - \omega^2 x'_D - \dot{\omega} y'_D\\ \ddot{y}_D = \ddot{y}_G - \omega^2 y'_D + \dot{\omega} x'_D = \frac{ \sum_{i} F_{y,i}^{ext} }{m} - \omega^2 y'_D + \dot{\omega} x'_D\\ I_G'\dot{\omega} = \sum_{i}M_{G,i}^{ext} \end{array} \right. $$

I omitted the derivation of these formulas as it's quite lengthy, but it's just rigid body motion and can easily be derived by taking the second time derivative of the position of the D point with respect to the fixed coordinate system.
We now make the assumption that angular velocity is zero at the beginning ($\omega=0$), as the stick is initially still, and that the corner won't lift from the ground or go below ground level ($\ddot{y}_D=0$) as these assumptions are true in both cases:

$$ \left\{ \begin{array}{c} \ddot{x}_D = \frac{ \sum_{i} F_{x,i}^{ext} }{m} - \dot{\omega} y'_D \\ 0 = \frac{ \sum_{i} F_{y,i}^{ext}}{m} + \dot{\omega} x'_D \\ I_G'\dot{\omega} = \sum_{i}M_{G,i}^{ext} \end{array} \right. $$

And here is where friction enters the game.

Case 1: friction force is strong enough

This equals to say we have as much friction force available as we want, or in a more mathy way $\mu_s\rightarrow\infty$. If this happens, we can assume that the corner will remain a fixed point and that the block will pivot about it ($\ddot{x}_D=0$):

$$ \left\{ \begin{array}{c} 0 = \frac{ \sum_{i} F_{x,i}^{ext} }{m} - \dot{\omega} y'_D \\ 0 = \frac{ \sum_{i} F_{y,i}^{ext}}{m} + \dot{\omega} x'_D \\ I_G'\dot{\omega} = \sum_{i}M_{G,i}^{ext} \end{array} \right. $$

At this point it's just a matter of filling in the variables and solving the system. We have 3 equations and 3 unknowns, namely $\omega$, $F_{friction}$ and $F_{normal}$. We also know that $y'_D = -\frac{L}{2}$ and $x'_D = -\frac{a}{2}$, where $L$ is the total length of the stick:

$$ \left\{ \begin{array}{c} 0 = F_1 + F_{friction} - F_2 + m \dot{\omega} \frac{L}{2}\\ 0 = F_{normal} - W - m \dot{\omega} \frac{a}{2}\\ I_G'\dot{\omega} = F_1 \left( \frac{L}{2}-a \right) + F_2 \left( 5a-\frac{L}{2} \right) +F_{friction} \frac{L}{2} - F_{normal} \frac{a}{2} \end{array} \right. $$

And now the system can be solved:

$$ \left\{ \begin{array}{c} F_{friction} = F_2-F_1 - \frac{ 3aL \left( 5F_2 - F_1 - \frac{W}{2} \right)}{2m(L^2+a^2)} \\ F_{normal} = W + \frac{3a^2 \left( 5F_2-F_1- \frac{W}{2} \right)}{2m(L^2+a^2)} \\ \dot{\omega} = \frac{3a^2 \left( 5F_2-F_1- \frac{W}{2} \right)}{2m(L^2+a^2)} \end{array} \right. $$

Where $F_1$ is the force pointing towards the right, $F_2$ is the force pointing towards the left, and $W$ is the weight of the stick. I also assumed that $I_G' = \frac{1}{12}m(L^2+a^2)$ as the stick is a rectangle. Doing this allows us to collect together some letters.

Case 2: friction force is not strong enough

In this case we just get back to this point:

$$ \left\{ \begin{array}{c} \ddot{x}_D = \frac{ \sum_{i} F_{x,i}^{ext} }{m} - \dot{\omega} y'_D \\ 0 = \frac{ \sum_{i} F_{y,i}^{ext}}{m} + \dot{\omega} x'_D \\ I_G'\dot{\omega} = \sum_{i}M_{G,i}^{ext} \end{array} \right. $$

as we cannot assume that the corner will be a static pivot for the body. In this case the path is very similar to before because we still have 3 unknonws, but this time we suppose that $F_{friction} = \mu_d F_{normal}$ because the corner is sliding and is subject to constant friction force:

$$ \left\{ \begin{array}{c} m \ddot{x}_D = F_1 + \mu_dF_{normal} - F_2 + m \dot{\omega} \frac{L}{2}\\ 0 = F_{normal} - W - m \dot{\omega} \frac{a}{2}\\ I_G'\dot{\omega} = F_1 \left( \frac{L}{2}-a \right) + F_2 \left( 5a-\frac{L}{2} \right) +F_{friction} \frac{L}{2} - F_{normal} \frac{a}{2} \end{array} \right. $$

which can again be solved:

$$ \left\{ \begin{array}{c} \ddot{x}_D = F_1+\mu_dF_{normal} - F_2 + \frac{(F_1-F_2+\mu_dW)3L^2 + (10F_2-2F_1-W)3aL}{L^2+4a^2-3\mu_daL} \\ F_{normal} = W + \frac{(F_1-F_2+\mu_dW)3aL + (10F_2-2F_1-W)3a^2}{L^2+4a^2-3\mu_daL} \\ \dot{\omega} = \frac{(F_1-F_2+\mu_dW)6L + (10F_2-2F_1-W)6a}{m(L^2+4a^2-3\mu_daL)} \end{array} \right. $$

Bonus case: no friction

For this you can jus take the previous solution and set $\mu_d = 0$ to get the solution where the stick slips completely:

$$ \left\{ \begin{array}{c} \ddot{x}_D = F_1 - F_2 + \frac{(F_1-F_2)3L^2 + (10F_2-2F_1-W)3aL}{L^2+4a^2} \\ F_{normal} = W + \frac{(F_1-F_2)3aL + (10F_2-2F_1-W)3a^2}{L^2+4a^2} \\ \dot{\omega} = \frac{(F_1-F_2)6L + (10F_2-2F_1-W)6a}{m(L^2+4a^2)} \end{array} \right. $$

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  • $\begingroup$ I started to share my new questions, new questions will come during the day, I'm doing a thesis work on static calculations of a tall building, the questions will continue to be posted for a day or two, thank you. You can see the question I asked on the link. engineering.stackexchange.com/questions/55612/… $\endgroup$ Jun 27, 2023 at 14:55

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