# How does the center of mass move forward on a rocking chair?

When I was rocking on a recliner at home I noticed the combined center of mass of me and the chair would move horizontally (also vertically).

The only apparent forces acting on this system is gravity and the normal force (on the chair), both of which cannot affect the motion of the center of mass horizontally. And the forces between me and the chair are internal.

How has the center of mass gone forward, what have I not factored in?

• There's also tangential friction between the floor and the chair Dec 17, 2020 at 12:00
• Both of which cannot affect the motion of the center of mass horizontally Oh yes they can, gravity produces torque about contact point. Why do you think drunkards falls down to the ground ? They loose balancing abilities to compensate gravity induced torque. Torque produces movement in both $x,y$ axis. Same principle here. Dec 17, 2020 at 12:08
• @AgniusVasiliauskas Strictly speaking the vertical reaction force of the floor and gravity don't affect the horizontal motion. According to the vectorial balance of momentum, forces along a specific direction cannot alter the momentum component along an orthogonal direction. Dec 17, 2020 at 12:38
• @pglpm. Ok then explain why drunkards falls to ground (has horizontal movement component too)? Would they fall down in an orbital space station ? Dec 17, 2020 at 13:30
• @AgniusVasiliauskas If a drunkard loses balance on a very slippery floor, the mass-centre will only move vertically. It's the horizontal friction that changes the horizontal momentum, not the vertical floor reaction or gravity. I can agree with you in the sense that the vertical forces affect the balance of rotational momentum, which in turn affects the horizontal friction. But then we can also agree, by the same logic, that the horizontal friction affects the vertical motion. Dec 17, 2020 at 14:41 I've replaced the rocking chair with a circle with a smaller inset circle representing the centre of gravity (CoG). The weight is $$mg$$.

Now assume there's plenty of friction between the circle (rocking chair) and floor, so as to prohibit any sliding/slipping.

At this point the distance between the vector $$\mathbf{mg}$$ and the vertical running through $$\text{O}$$ is $$L$$.

Gravity now forces the CoG to be lowered. A torque $$\tau$$ about the point $$\text{O}$$ arises:

$$\tau=mgL$$

This torque causes rotational acceleration of the circle:

$$\tau=I\alpha$$

This angular acceleration causes not only to move the CoG downward but also to the right.

Note that due to inertia the system, unimpeded, will enter into an oscillatory mode (hence 'rocking chair')

• Diagram would be more like a rocking chair with the CM in a lower quadrant--in my experience, if you try to rock $2 \pi / 3$ from equilibrium, you fall over :). Dec 17, 2020 at 14:36

How has the center of mass gone forward, what have I not factored in?

You have not factored in the horizontal force of friction at the ground.

• So are you saying one could not start rocking if the chair were on a perfectly frictionless surface (as an approximation, imagine wet ice)? Interesting, but I'm not sure it's true. Dec 18, 2020 at 8:54
• It's true. The overall center of mass of the chair and sitter combinded cannot start to move horizontally unless there's some horizontal (i.e. not purely vertical) interaction with the surface or some other external object. Maybe it could start rocking but as the top of the chair moved foward the base would have to slide back to keep the COM in one spot.
– bdsl
Dec 18, 2020 at 9:34
• Try a rocking chair on a carpet on a smooth floor… the center of mass won't move, the carpet will. Dec 18, 2020 at 9:39
• @Peter-ReinstateMonica said “So are you saying one could not start rocking if the chair were on a perfectly frictionless surface”. No. I am not saying that. I am saying that on a perfectly frictionless surface your center of gravity would not move horizontally. Instead the point of contact would slip back and forth as you rocked. Bergi and bdsl have the right idea
– Dale
Dec 18, 2020 at 12:44
• @Dale Well yeah, that's not the kind of rocking commonly meant with the word, where your center of mass goes back and forth (and you can continue for a while with minimal movement relative to the chair). Dec 18, 2020 at 15:31

If you're just curious about why the instantaneous system of forces has a horizontal component which leads to horizontal motion (w.r.t. the earth), then you can consider that the system is instantaneously similar to an inverted pendulum. If the friction doesn't allow any slip between floor and rocking chair, then the centre of mass is instantaneously constrained to move on a tangential motion around the pivot on the floor, as in the following diagram: The pivot is the black dot; the centre of mass is the blue dot; its constrained instantaneous direction of motion is the dashed red curve.

This situation implies that the reaction force of the floor is directed as the green vector $$\pmb{F}_{\text{R}}$$ in the picture, and its magnitude must balance the component of the gravitational force $$\pmb{F}_{g} \equiv m\pmb{g}$$ along the line connecting centre of mass and pivot. With some trigonometry we find that this magnitude is $$\lvert \pmb{F}_{\text{R}} \rvert = \lvert \pmb{F}_g\rvert\ \cos\theta$$, in terms of the angle $$\theta$$ depicted in the picture.

So the total force on the centre of mass is $$\pmb{F}_{\text{T}} = \pmb{F}_{g} + \pmb{F}_{\text{R}}$$, and again with some trigonometry we see that by construction it must be tangential to the dashed red curve. It therefore has a vertical component and a horizontal component: If there is no friction between floor and chair (say because of a layer of oil), then the reaction force would be purely vertical upwards, and its magnitude would be less than $$F_g$$, because it would only compensate for the weight of the part of the system directly above it. Then the total force would also be purely vertical and directed downwards, and the centre of mass would not have any horizontal motion.

This latter case is more or less what happens when people slip and fall on the floor: typically their feet go horizontally one direction, say backwards, and the face goes the opposite direction, say forwards. And the centre of mass falls more or less vertically.

It may be worth noting that in this kind of problems our deductions using Newton's laws sometimes go in the direction "motion $$\rightarrow$$ forces", rather than "forces $$\rightarrow$$ motion": we experimentally see that there is horizontal motion and no slip on the floor, and therefore conclude that there must be horizontal friction. This was exactly your kind of deduction :)

With all the brilliant answers I received, I think I can simplify the problem.

We can replace the recliner with a rod tending to slip on a surface. As in both cases, the center of mass moves forward which is my main query.

If the surface was frictionless the rod will fall in such a way that the center of mass falls vertically downward.If friction was sufficient to avoid slipping the center of mass would pivot around the bottom point of the rod just like Gert and pglpm said.

Friction is the factor I forgot to take.