# 4 newtons of force???

The rules say that the mini bot has to push the sensor with a force of 2-4 newtons. I think that it would be a good idea to try to go for a force of 4 newtons but I am curious as to how much force 4 newtons is.

What are some examples of how much force this is and what mass and velocity a bot would have to be moving to create such a force? Is this accomplishable by the “spring loaded” bots that people are talking about?

.899 lbs of force, idk specific examples

4 Newtons is about .9 of a lb. So a fairly strong force for that small of a robot, but I am sure you can get that much of a force.

If you have a mass of 2kg, then this means you need an acceleration of 2m/s^2

Now, you assume that you are working with a free floating object (no friction/gravity), and approximately 1/4" or 0.0063m of stopping distance (based on the rules).

This means, that you need to solve for u, or initial velocity in this formula.

v=final velocity=0
u=initial velocity=unknown
a=acceleration=2
s=distance traveled=0.0063

v^2 = u^2 + 2as

Based on that I am coming up with about 0.16m/s
Now, this is actually low, due to the fact that the calculation I just did is ignoring gravity.

I am using a formula from
http://tutor4physics.com/motion1d.htm
So, I dunno if it is correct, but it sounds about right. Admittedly, I need to brush up on Newtonian physics

Physics Content:
1 Newton=0.225 lbs
As such, one way to trigger the plates is to push upwards with ~.9 lbs of force.

A robot moving upwards has some momentum § defined as the robot’s mass multiplied by its velocity:
P=mv
Assume that when the robot hits the button it exerts a constant force F until coming to a stop. Using Newton’s second law:
F
delta t=P where delta t is the time required for the robot to come to a complete stop (v=0, P=0).
or
F=P/(delta t)

As such, a robot moving fast (high P) could trigger the button even if it could not produce the necessary force when stationary.

so basically we have found that it does not take a lot of speed to generate 4 newtons of force with a 4.5lb mini bot?

This is a lot of science content and I don’t think that i comprehend all of it seeing that I am still in high school and i am currently in week 2 of my physics class

What I showed you is one dimensional Newtonian mechanics. You will learn it in your physics course. (probably)

i will try to make sense.
notes:
110" = 2.794meters
1/4" = .00635 meters
g=9.8 is the acceleration of gravity
m=mass, v=velocity, h=height from base
Fd=work done (force x distance)
(i.e. 4Newtons at top of the pole x .00635 meters to move plate)

Energy is always conserved. E = E
0.5mv^2 + mgh + Fd = 0.5mv^2 + mgh + Fd
(at bottom of the pole) = (at top of the pole)

0.5mv^2 + 0 (h=0) + 0 (no work done)=
0 (v=0) + m(9.8)(2.794meters) + (4Newtons)(.00635meters)

.5mv^2 = 27.3812m + .0254

mv^2 = 54.7624m + .0508

v^2 = 54.7624 + (.0508/m)

v = sqrt(54.7624 + (.0508/m))

this looks complicated: however, m (mass) is in kilograms.
Unless your bot is less than .1kg = .22lbs, that number will be negligible.

v approximates sqrt(54.7624) =7.4… m/s, i’m going to round up to 7.5 to be safe.

you need MINIMUM 7.5 m/s = 25 fps upward launching velocity if you were to launch an object from the base that does not touch the pole but effectively trips the sensor. (like a cylinder ring) This accounts for gravity not for air resistance (low) and friction (rubbing against the pole), if you had friction you should focus on programming a robot that actually climbs rather than launches.

concerning a push from a moving/climbing/(or launched) robot,

the energy transferred to the sensor is = Kinetic Energy (due to motion)
E = 1/2 m v^2 = Fd = (4N)(.00635meters) =.0254 Joules