Figure1 attached shows the top view of a vehicle with standard wheels
attempting to rotate counter-clockwise but not yet moving because the
torque being applied to the wheels is too low. The +Y axis is the
Assume the following:
Four identical standard non-steerable wheels of radius “r”
Wheels are located at the corners of a rectangle
Axis of each wheel is parallel to the X-axis
Coefficient of friction “mu” is the same in all directions
The same magnitude torque “tau” is being applied to each wheel
Let f2 = trackwidth/wheelbase
The right wheels are being torqued “forward”
The left wheels are being torqued “backward”
The vehicle is in static equilibrium
CoM aft of CoG. Vehicle is on a flat, level floor.
Let f1 = (distance from CoG to CoM)/(distance from CoG to the point midway between rear wheels)
Let W be the weight of the vehicle.
Fn is the total friction reaction force of the floor on the bottom of wheel #n,
and Fnx Fny are its components. n= 1,2,3,4.
Find the torque tau in terms of mu, r, W, f1, and f2 required to break the
static friction and start the vehicle rotating.
The Center of Mass (CoM) being located aft of the Center of Geometry (CoG) affects the front/rear weight distribution: the Normal force on the front wheels is less than the Normal force on the rear wheels.
This in turn makes the maximum available static friction at the front wheels less than the rear wheels. The analysis of the actual static friction forces* at the wheels, as long as they are less than the maximum available static friction, is unaffected by the weight distribution.
i.e. the Fn force (with Fnx Fny components) at each wheel
Solving each wheel generally for static equilibrium (from quiz 6):
Distance from wheel to CoG = sqrt(trackwidth^2+wheelbase^2)/2 and from now on is d
Fnydsin(angle included) = Fnxdsin(other angle included) because net torque is zero.
Fny(trackwidth/d) = Fnx(wheelbase/d)
Fnx = Fnyf = Tauf2/r
|Fn| = |Fnx + Fny| = Tausqrt(f2^2 + 1)/r = Nnmu
I think we only have to exceed the smallest static friction force (F1 and F2) to create a net torque on the bot so:
Tausqrt(f2^2 + 1)/r > N1mu
Tau > N1mur/sqrt(f2^2 + 1)
We can solve the forces in the yz plane for one side of the robot:
let wheelbase = b and distance from CoM to CoG = x
f1 = 2x/b
Tau about CoM (in the yz plane) is zero, therefore N1(b/2+x) = N4(b/2-x)
Divide by b/2, so N1(1+f1) = N4(1-f1)
Fnet is also zero, so W/2 = N1+N4
N1 = (W/2-N1)(1-f1)/(1+f1)
N1 = (W/4)(1-f1)
Now we can solve Tau in terms of givens:
Tau > (W/4)*(1-f1)mur/sqrt(f2^2 + 1)
[strike]But I think I’m making the wrong assumption in the second section. If we have to exceed the greater static friction, Tau > (W/4)*(1+f1)mur/sqrt(f2^2 + 1). I’m not so sure how the free-body diagram works out. On the other hand, I could be totally wrong altogether…[/strike] (Thanks for the solution, ether!)