Friction generated from the rotation of the wheels provides a torque to slow down their rate of rotation. As the entire chair slows down, friction between your body and the seat of the chair slows *you* down. That's the only major contributor.

A phenomenological approach would be to assume the wheel friction provides a force to slow the chair down proportional to $v^{\alpha} m^\beta$ where $v \equiv dx/dt$ is the velocity of the chair, $m$ is the mass of the Argus + chair system, and both $\alpha$ and $\beta$ are unknown exponents. Let $K$ represent the unknown proportionality constant. Then

$\begin{equation}
m v \frac{dv}{dx} = - K v^{\alpha} m^\beta
\end{equation}$

Integrating this equation up to the location $d$ where $v = 0$ yields

$\begin{equation}
d = \left(\frac{1}{2 - \alpha} \right) \left (\frac{1}{K} \right) {m^{1 - \beta}} {v_i}^{2 - \alpha}
\end{equation}$

where $v_i$ is the speed you start at from your initial push. You can see that if $\alpha < 2$ then you'll stop after traveling a longer distance if you start out faster. When $\alpha > 2$ you never actually come to rest according to this model, but at a given time you'll always be farther along if you started with a higher velocity.

I have not yet considered the complications arising from rotational motion of your chair if you push off at an angle. That is left as an exercise to the reader.

If you choose to perform an experiment, please take proper safety precautions and consider wearing a helmet. Also assess what sort of damage you might cause to the floor and what sort of budget you must set aside to repair it (or how to successfully avoid responsibility for doing so). Or, stop playing with your chair and get back to what you're supposed to be doing :-)

If you neglect friction, the strap presses on the cylinder, and exerts a force perpendicular to the contact surface. So the infinitesimal length of strap sitting on the highest point of the cylinder exerts force downward, but at any other point, there will be a horizontal component as well. If the configuration of your strap is symmetrical, any horizontal force exerted on one side of the cylinder will be compensated by an equal, opposite force exerted on the symmetrical side.

So the overall force exerted on the cylinder has no horizontal component, only vertical, and equal to $2T \sin \alpha$, where $T$ is the tension of the strap, and $\alpha$ the angle the ends of the strap make with the horizontal. This is also the reaction that the truck bed will exert on the cylinder from below.

## Best Answer

There are two effects at work there, as follows.

First, buckets like that are injection-molded, which means they are formed with a slight "draft angle" or taper to their sides, which makes it easier to get them out of the mold in which they were formed. The wall thickness is typically constant throughout the bucket, which means the outside taper angle matches exactly the inside taper angle. What this means is that if you scrunch one bucket down inside another bucket, they fit together extremely snugly from a geometry standpoint, and the slight elasticity of the bucket material causes the inside bucket to contract slightly and the outside bucket to stretch slightly. All these effects create a bucket pair which, when firmly scrunched together, tend to cling extremely tightly to one another via mutual friction.

The second effect is that to pull them apart, air needs to find its way into the gap between the two buckets. Because of how perfectly they fit together, there is essentially no gap between them and hence almost no way to get any air to flow down into the space between them.

These two effects guarantee that it is almost impossible to pull a pair of identical plastic buckets apart once they have been pressed together.