Propulsion that does not involve driving wheels

[comphappy]

Originally Posted by EricH

Frictional force (static) = Ffs
Max frictional force = mu*N, N = the normal force. So far so good, right?

NO
Lets lay down some facts
is u of static friction which FIRST has at 0.06 for static inline.
the Fn is going to be mg asumming the robot is flat on the ground.
120 lb = 54.43 kg
Fn = 54.43 * g
Fn = 533.41 N
now lets calculate the maximum frictional force that can be exerted by wheels before they slip on the surface (static friction as the wheel does not "leave the surface")
Fu = Fn * u
Fu = 533.41N * 0.06 = 32N
now lets find out the maximum acceleration because we must be equal to or less then Fu.
Fu = ma where a is the acceleration parallel to the surface
32 = 54.43 * a
a = 0.58 m/s/s

low lets try that with a robot at 10kg

Fn = 10 * g =98N
Fu = 98N * 0.06 = 5.88 N
5.8 = 10 * a
a = 0.58

Now Lets try it with the equation that I derived
a=u*g
a = 0.06 * 9.8 = 0.58 m/s/s
So with no wind propulsion which I said in my first response. mass does not mater in terms of acceleration.

N = m*g, where m = the mass of the object (robot) and g = either 9.8 m/s^2 or 32 f/s^2 depending on your system of measurement. Am I not correct?

For the condition where nothing is slipping or about to slip, Ffs <= mu*N. When the object is about to slip, Ffs = mu*N. When the object slips, you get into Ffk, or the force of kinetic friction. I'm still correct in this, right?

No the m is N you are putting that in twice, which is wrong, see the image that I slipped in my last post.

By your equations, you are assuming that the wheel is about to slip. That is, F=m*a (the standard force equation) == mu*N. But wait! Where, oh where, does Ffs come in? It is the force exerted by the static friction, so it is m*a also, I'm assuming. Please correct me if I am wrong here.

This is not right, see above.

Now, on to business.

Ffs = m*a = mu*m*g, assuming a flat plane. I think this is quite reasonable considering the application and that you're about to slip your wheels.

So, m cancels. a = mu*g. You are correct. The mass does not matter.

mass is not in there twice take it out.

HOWEVER: you add a fan using your available mass (which doesn't matter) which adds additional force going downwards, correct? mu is constant due to the rules, so we can set that aside. That leaves g and a. a = g. g is a component of the normal force, under N = m*g. The normal force is equal and opposite to the weight m*Ag , such that W (weight) + N = 0. So far so good, right?

When you add the force Ffan going downwards (or upwards), you now have W + N + Ffan = 0. Ffan would be added to either W or N if the fan were pushing down, and you have W+Ffan = -N or N + Ffan = -W.

Your full new equation, therefore, would read: m*a = mu*m*g + Ffan. (Frictional Force = Normal force times mu plus Fan force) Correct me if I am wrong here.

Fan force is dependent on area and speed, is it not? There is not a mass component there, because it is already accounted for. So you can no longer divide out mass. (basic algebra)

As for your other comment, YOU also need to watch your equations. Unless you can point out what is wrong with my response to your example of a 500 lb and a 10 lb robot, that stands. I could also run it in metric if I wanted to.

I have pointed it out, in metric (not sure why that mattered). As for above with a fan you can accelerate a little more. although the equations above for the fan suffer from the same error as the wheels.

This is important to make clear as intuition for most people is wrong here.