Physics Quiz 7

Technical Technical Discussion
#1

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
“forward” direction.

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.

Problem:

Find the torque tau in terms of mu, r, W, f1, and f2 required to break the
static friction and start the vehicle rotating.

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Figure 1.png
342bf8418cac3962334d8efd47740eb080cfdf39.png859×789 4.1 KB



Figure 1.png
342bf8418cac3962334d8efd47740eb080cfdf39.png859×789 4.1 KB

#2

*Hint:

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

#3

It’s the holidayyyys, ether, we can’t do physics until after New Year’s day!

I’ll work on it tomorrow morning, I just forgot it was here.

#4

Ether,

This question seems to be a bunch of work meanwhile, it does not seem to get any interesting concepts across.

I read over it and it did not give me much inspiration to solve it.

#5

I assumed that the CoG is halfway between the front and back, if that is false pretty much none of this makes sense.

Fn(max) = Nn*mu

Since its symmetrical:
N1 = N2
N3 = N4

summing forces
W = N1+N2+N3+N4 = 2N1 + 2N3
summing moments about axis of front axles, defining wheelbase as 2 units
W*(1+f1) = (N3+N4)2 = 4 * N3
combining formulas
W= 2N1 + W*(1+f1)/2
W(1-1/2-(f1)/2) = 2*N1
N1 = W(1-f1)/4

F1(max) = F2(max) = muW(1-f1)/4
F3(max) = F4(max) = muW(1+f2)/4

summing moments around vertical axis through CoG
0=F1y1+F2y1+F3y1+F4y1-F1xf2-F2xf2-F3xf2-F4xf2
( F1x + F3x )*f2 = F1y + F3y

by Pythagorean theorem:
(sqrt((F1)^2-(F1y)^2) + sqrt((F3)^2-(F3y)^2))*f2 = F1y + F3y

Fny = tau/r

(sqrt((muW(1-f1)/4)^2-(tau/r)^2) + sqrt((muW(1+f1)/4)^2-(tau/r)^2)) f2 = 2tau/r

I spent around half an hour trying to simplify that but just made it worse:

sqrt((muW(1-f1)/4)^2-(tau/r)^2) * f2 = 2tau/r - sqrt((mu*W(1+f1)/4)^2-(tau/r)^2)) *f2

(muW(1-f1)/4)^2-(tau/r)^2) * (f2)^2 = 4 * (tau/r)^2 - 4 * tau/r * sqrt((muW(1+f1)/4)^2-(tau/r)^2)) f2 + ((muW(1+f1)/4)^2-(tau/r)^2)*(f2)^2

4tau/r * sqrt((muW(1+f1)/4)^2-(tau/r)^2)) f2 = 4(tau/r)^2 + (f2)^2 * (((muW)^2f1)/4)

16 * (tau/r)^2 * ((muW(1+f1)/4)^2-(tau/r)^2) (f2)^2 = 16 * (tau/r)^4 + 8(tau/r)^2 * (f2)^2 * (((muW)^2f1)/4) + (f2)^4 * (muW)^4 * ((f1)^2)/16

#6

It is not a bunch of work; it is rather straightforward to solve. That is the interesting concept.

#7

*Nice try. See note embedded below in red.

#8

Solving each wheel generally for static equilibrium (from quiz 6):

  1. Fny=Tau/r
  2. Distance from wheel to CoG = sqrt(trackwidth^2+wheelbase^2)/2 and from now on is d
  3. Fnydsin(angle included) = Fnxdsin(other angle included) because net torque is zero.
  4. Fny(trackwidth/d) = Fnx(wheelbase/d)
  5. Fnx = Fnyf = Tauf2/r
  6. |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:

  1. Tausqrt(f2^2 + 1)/r > N1mu
  2. Tau > N1mur/sqrt(f2^2 + 1)

We can solve the forces in the yz plane for one side of the robot:

  1. let wheelbase = b and distance from CoM to CoG = x
  2. f1 = 2x/b
  3. Tau about CoM (in the yz plane) is zero, therefore N1(b/2+x) = N4(b/2-x)
  4. Divide by b/2, so N1(1+f1) = N4(1-f1)
  5. Fnet is also zero, so W/2 = N1+N4
  6. N1 = (W/2-N1)(1-f1)/(1+f1)
  7. N1 = (W/4)(1-f1)

Now we can solve Tau in terms of givens:

  1. 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!)

#9

You got it.

Solution attached.

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solution.pdf (35.5 KB)


solution.pdf (35.5 KB)