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  1. What is slowing me down when I push off my desk?
  2. What are the "Major" contributors? and
  3. Does how hard I push with one hand (or with two hands) make a significant difference?
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@Global nomad: looks much better that way any ideas? –  Argus May 28 '12 at 22:20
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It's friction in the wheels, the air friction is probably negligible. Why is there controversy? For the third question, make a significant difference to what? –  Ron Maimon May 28 '12 at 22:39
    
make a significant difference in my goal of pushing myself to the minifridge :) –  Argus May 28 '12 at 22:48
    
There's friction in the wheels, which is probably constant, so will probably slow you down by a certain amount of speed per second. So your position $x$ as a function of time $t$ will be parabolic decreasing to $0$. So you need to push hard enough to get sufficient initial velocity. One hand or two - up to you. –  Mike Dunlavey May 29 '12 at 0:08

1 Answer 1

up vote 4 down vote accepted

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 :-)

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"You can see that if alpha<2 then you'll go farther if you start out faster, but if alpha>2 then you'll go farther if you start out slower." You sure about that? I don't think it fits with your initial definition of alpha. –  Colin K May 29 '12 at 2:29
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@ColinK, you're right, I made a dumb mistake in my first comment on what happens when $\alpha > 2$. I'm fixing the answer now. Thanks. –  kleingordon May 29 '12 at 3:44
    
I wouldn't call it dumb, and anyway it's fixed now. +1 from me :) –  Colin K May 29 '12 at 15:47
    
+1 a very good answer. This will help with what safety equipment is necessary to overcome risk of injury. Is there a standard force I could establish to push. How would I get a consistant push if multiple attempts are made to proove the Mini-fridge is not too far away. –  Argus May 29 '12 at 22:21

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