Chief Delphi - Physics Quiz 5

Chief Delphi (http://www.chiefdelphi.com/forums/index.php)

- Technical Discussion (http://www.chiefdelphi.com/forums/forumdisplay.php?f=22)

- - Physics Quiz 5 (http://www.chiefdelphi.com/forums/showthread.php?t=98947)

See attached PDF "Problem Statement".

Page 1 is a diagram.

Page 2 has the assumptions and problem statement.

Re: Physics Quiz 5

I'm having trouble visualizing the wheel orientation.

Re: Physics Quiz 5

Quote:

Originally Posted by DonRotolo (Post 1092578)

I'm having trouble visualizing the wheel orientation.

It's a standard-wheel vehicle. The wheels are not steerable. All four wheels are oriented "forward" (the front of the vehicle is labeled in the diagram).

Re: Physics Quiz 5

No, not quite that easy. But not too much harder. The vehicle is in static equilibrium. The wheels are not accelerating. Therefore, the net torque on each wheel (in the plane of the wheel) must be zero. Your solution does not satisfy this physical requirement.

Quote:

Originally Posted by quinxorin (Post 1092591)

It can't be this easy.
A)
Fy=Fx=(T*sqrt(2))/(2r)
and
F=T/r
B)
T=μNr

Work:
Part A:
T=Fr
(F is the same as force applied by vehicle)
T/r=F
Because the wheels are in a square, theta (here "x") is 45degrees:
Fy=Fx=Fcos45=Fsin45=F(sqrt(2)/2)

Part B:
F=μN
T=FR
T=μNr

Re: Physics Quiz 5

It would appear that friction is trying to crush your robot into a diamond!

Re: Physics Quiz 5

I know I'm doing something wrong.

1. The wheel applies a force of Tau/R in the y direction. There is no motion in the wheel so Fy must equal Tau/R.
2. There is no motion of the robot, so there must be a torque to resist that generated by Fy, namely that generated by Fx. Fx and Fy are at the same distance from the CoM and at opposite angles to the moment arm, so they must be equal in magnitude.
3. F is the magnitude of Fy added to Fx: Tau*sqrt(2)/R
4. To rotate, Fy must exceed Fx. This yields a net torque in the clockwise direction, which is what I assume is the goal.
5. Fy>Fx
6. Tau/R>muN
7. Tau>muNR

I have a problem with my answer for B because I feel like Tau/R should not exceed the maximum static friction, because then there would be a lot of wheel spin, which doesn't click for me. I also realize it's what quinxorin already posted. I just don't know how to do it otherwise.

Re: Physics Quiz 5

See notes in red below:

Quote:

Originally Posted by Ninja_Bait (Post 1092624)

I know I'm doing something wrong.

1. The wheel applies a force of Tau/R in the y direction. the component of the floor's reaction force on the wheel in the Y direction is tau/r There is no motion in the wheel so Fy must equal Tau/R. correct
2. There is no motion of the robot, so there must be a torque to resist that generated by Fy, namely that generated by Fx. Fx and Fy are at the same distance from the CoM and at opposite angles to the moment arm, so they must be equal in magnitude. correct
3. F is the magnitude of Fy added to Fx: I assume you mean vector addition. Tau*sqrt(2)/R correct

Here's where you go off the rails:
4. To rotate, Fy must exceed Fx. This yields a net torque in the clockwise direction, which is what I assume is the goal.
5. Fy>Fx
6. Tau/R>muN
7. Tau>muNR

I have a problem with my answer for B because I feel like Tau/R should not exceed the maximum static friction, because then there would be a lot of wheel spin, which doesn't click for me. I also realize it's what quinxorin already posted. I just don't know how to do it otherwise.

Re: Physics Quiz 5

Yeah, that's what I figured. :\

Let me try again:
At equilibrium, F=muN
Tau*sqrt(2)/R=muN
Tau=muNR/sqrt(2)
If Tau is any greater, it will overcome F?

Re: Physics Quiz 5

Notes in red below:

Quote:

Originally Posted by Ninja_Bait (Post 1092634)

Yeah, that's what I figured. :\

Let me try again:
At equilibrium, at the "edge" of equilibrium, just before things break loose F=muN correct
Tau*sqrt(2)/R=muN correct
Tau=muNR/sqrt(2) correct
If Tau is any greater, it will overcome F? correct!

Re: Physics Quiz 5

YAY! I can do physics!

Re: Physics Quiz 5