# Predicting Rotation vs Sliding Forwards when Pushing an Object

Suppose there is a cuboid object sitting on a table, and a force is then applied to it. This figure shows the top-down view:

I am curious to understand how the object will move due to this force, such as whether it will rotate, or slide forwards. I would like to know how this will differ when the only varying parameter is the frictional coefficient between the object and table.

After pushing various objects around on my desk, I have noticed three different behaviours for different objects:

Sliding forwards:

My understanding is that the object will move according to whichever motion requires the least force, i.e. whichever motion would happen "first" if the force started from zero, and started to increase (e.g. from here). Given this, it seems to me that for a fixed value frictional coefficient, the motion requiring the least force would be the one where the object rotates about the end. My reasoning is that this motion requires the least physical movement of the object from its original position, i.e. the smallest integral of (force X distance) across the object, and therefore the least energy required to overcome the friction during sliding.

However, this does not seem to be the case in practice. I observe all three motions for different kinds of objects, and I don't understand why. Can somebody explain to me why my intuition is incorrect, and what the correct relationship is between the coefficient of friction, and the object motion?

• Do you find the same thing happens for the same objects? Or do the same objects undergo different scenarios? My first inclination is to say that it really just depends on where static friction "fails" first, which could have somewhat of a "stochasticity" to it based on irregularities in the surfaces that you cannot control. Nov 19, 2018 at 13:06
• The question and my answer here somewhat relates to this question, although it is probably more of an ideal scenario physics.stackexchange.com/a/435451/179151 Nov 19, 2018 at 13:10
• There doesn't seem to be much stochasticity in the experiments I've been doing. If I push the same object in the same place, it moves in the same way. But then if I push a different object of a similar size in a similar place, it can behave very differently. Nov 19, 2018 at 13:11
• It could still depend on irregularities in the surface of each object. Nov 19, 2018 at 13:15
• @Karnivaurus By conducting the experiment with different objects I believe too many variables are introduced to justify comparing the three different results. Lets take "flatness" to name just one. If the objects are not perfectly flat the normal forces that determine friction will vary over the surface. For example, an object may rotate about the center due to the object being bent concave upward, so that the maximum normal force occurs near the center with little or no friction resistance near the ends. Nov 20, 2018 at 14:15

For the ideal scenario, the path the cuboid takes is independent of friction. The primary way your desk experiments differ from the ideal scenario is your assumption that the applied force is constant (namely in direction).

Let's lay out some assumptions for the ideal scenario:

1. The cuboid is uniform in mass.
2. The cuboid has a uniform coefficient of friction (no 'sticky' spots).
3. The force applied to the cuboid is truly constant.

There is a common assumption about frictional forces as well:

1. Frictional forces are always opposite in direction to motion (i.e. $$\hat f_k = - \hat v$$ )

### Claim 1:

Under assumptions 1 and 3 above, the cuboids path will be a combination of rotation about the center and translation of the center.

With a single force $$\vec F$$ applied to the cuboid, Newton's 2nd law will non-zero acceleration of the center of mass and non-zero rotational acceleration about the center of mass (provided $$\vec F$$ is offset from the CM).

$$\begin{array}{|l|cc|} \hline & \text{Translation} & \text{Rotation} \\ \hline \text{No Friction} & \vec F = m \vec a & \vec R \times \vec F = I \vec \alpha \\ \hline \end{array}$$

### Claim 2:

Under assumption 2 and 4 above, the path the cuboid takes is independent of friction.

$$\begin{array}{|l|cc|} \hline & \text{Translation} & \text{Rotation} \\ \hline \text{Friction} & \vec F + \vec f_k = m \vec a & \vec R \times \vec F + \vec \tau_k= I \vec \alpha \\ \hline \end{array}$$

This can be understood intuitively from assumption 4: if friction is always a drag force (i.e. directed behind you) it never has a component to the left or right. Consequently, it only speeds you up or slows you down; it doesn't change the path you take. A mathematical proof is given below.
If friction can't change the path the cuboid takes, it still does the same amount of rotation and translation regardless of the coefficient of friction.

### Problems with real world examples

Assumption 3 is the weakest assumption when applied to experiments. Take for example pushing a book's corner with a finger. In that case, the book likely only experiences a contact (a.k.a Normal) force perpendicular to the face:

To truly apply a constant force, you would need some sort of linkage. For instance, a push-rod that connects to a pin in the cuboid. Otherwise you are missing out on part of the applied force parallel to the face.
Without that linkage setup, the force will start performing translation of the cuboid, but then predominantly rotate as you get further into the path.

### Proof frictional forces don't change the path an object takes:

Begin with Newton's law, but break the forces up into components parallel to the velocity and perpendicular to the velocity: $$\Sigma \vec F_{||v} + \Sigma \vec F_{\perp v} = m \frac{d \vec v}{dt}$$ Let velocity be written as $$\vec v = v \hat v$$ (i.e. magnitude and direction). Then: $$\frac{d \vec v}{dt} = \frac{dv}{dt} \hat v + v \frac{d\hat v}{dt}$$ Applying the expansion of $$\frac{d\vec v}{dt}$$ and taking the dot product with $$\hat v$$: $$\Sigma \vec F_{||v} \cdot \hat v = m \frac{dv}{dt}\hat v \cdot \hat v + mv\frac{d \hat v}{dt} \cdot \hat v$$ The dot product $$\vec F_{\perp v} \cdot \hat v = 0$$ by definition, while $$\frac{d \hat v}{dt} \cdot \hat v = 0$$ is true because $$\hat v \cdot \hat v = 1$$ by definition. Thus: $$\Sigma F_{||v} = m \frac{dv}{dt}$$ Shows that Forces parallel to v can only change the magnitude of the velocity, not it's direction.