### Force and the acceleration of a RipStik

I decided to combine the concept of using weights on a rope from a couple years ago with the pulling concept of last year to try to again find an equivalency between the acceleration generated by a force and the acceleration of the RipStik.

I opted to use six kilograms which proved prescient, four would have been too little to overcome the massive friction with the beam above. I need a ladder, aluminum foil, and maybe light grease to make that rope run free. A clamp on pulley would be even better. Duck tape to keep the weights safely on the hangar is also a future need.

I used the foam core radiant energy reflectors as cushioning.

The basic layout can be seen above. I should have positioned the weight farther south, to the left, as I had to hold a post on the south side of the walk at start. This caused the rope to "walk" south as I rode forward, which put the weights to the south side of the pad.

The distance over which I would accelerate is equal to the fall distance of the 6.05 kg mass (three two kilogram masses plus a 50 gram hangar), 2.6 m. Lerina looks on while holding a portion of the tape measure.

The masses being lifted

The timer was told to time the fall of the weights. The logic was that my acceleration on the RipStik would last only as long as the weights were falling. Once the weights hit the ground, no more acceleration. The distance over which this happened on the RipStik would be the same as the distance for the fall.

F = mg: F = mweightg

The force that the weights provided is equal to the mass of the weights times the acceleration of gravity. The weights provided a force F equal to 6.05 kg * 9.79 m/s² = 59.2 Newtons of force.

F = ma: F = mdanaa

That 52.9 Newton force accelerated the 70 kg sum of Dana (65 kg) plus the RipStik (5 kg). Rearranging Force = mass × acceleration one obtains acceleration = force/mass. Using a = F/m, a force of 52.9 N should produce an acceleration a = 52.9/70 =  0.85 m/s²

d = ½at²

That force acted over a distance of 2.6 meters in a time of 4.44 seconds. Using d = ½at² to obtain a = 2d/t² = 2*2.6/4.44² = 0.26 m/s². That represents a whopping 69% loss of theoretically available acceleration, but then the rope is rubbing directly against the wood - the coefficient of static friction was particularly high. If I could get a piece of foil up there, that might help. Maybe even some light grease.

Post-run note that the masses have landed to the north of the intended landing pad. The launch was wobbly at best, I started to lose balance and rather unintentionally pulled on the rope to give myself an initial velocity. The RipStik can be seen in the background.

In the image above, I make calculations on the system. I had projected a 50% loss due to friction, the 69% loss exceeded my own predictions. To an order of magnitude, the force generated the expected acceleration. My thanks to Vanity for the excellent photographs - she captured the system in good detail.