I need to give a lesson on stability tonight, but it has been a while since I’ve done freebody diagrams.
Can you folks help idiot check me before I fill the students’ heads with lies?
For BEST, they have made what is essentially a rectangle on PVC pipe casters - see beautiful figure 1. I’m going to model it as a point mass with two struts. They push it to move it, and it has a tendancy to tip over. I want to create a simple formula to determine whether a redesign will tip, before they actually design it.
To do this, I’m calculating torques about the far strut. I see 3 contributors:
Fd1 = The force imparted by pushing times the height of the push point.
Nd2 = The normal force (weight) times the horizontal distance from the strut to the center of mass.
0 = The friction force from sliding over the carpet. I’m not so sure on this one - I think it is zero because it is right at the tip point?
This gives me Fd1 < Nd2 to ensure static stability. However, F is limited by friction, so I limit it to Nu.
This gives me Nud1 < Nd2, which becomes d2/d1 > u
This simple formula is pretty grokable, but is it true? Or more specifically, is it true enough?
Thanks guys, and be nice… our mentors are all ECEs, marketers and psychologists.
Correct. The distance is 0, so r * F will always be 0 for the torque due to friction around that point. This is the reason to choose the contact point of the leg.
Assuming that the device is in a static configuration, i.e. it’s not accelerating noticeably, all your math looks good. If you want to deal with acceleration, choose the center of mass of the device, and do torque calculations around that. Assuming that I’m not forgetting something, if you push below the center of mass, your equation is conservative when accelerating and should always return an answer that will work.
u, or μ as I prefer to denote it, is zero due to the roller, so you are correct that that force is zero. As for the torque part, that is also true, if I recall correctly–but a zero force has no torque application anyway.
The equations are a little trickier. The first one is correct–Fd1 < Nd2 for stability. However, making F=N*μ as a maximum probably isn’t the way to go here.
Let’s try this: Fd1=Nd2 for the point at which slip stops and tip starts. If N is known, and d2 is known, then F and d1 are unknown. d1 is the height of the force, which may be knowable. (For simple stuff, CG is the point of application.) That leaves F as unknown, and that means that Fmax = N*d2/d1.
[Note: The above assumes that it isn’t tipping because it hits something at that foot–if that were the case, you’d add a moment at the point, and Fd1 would need to be less than Nd2+the moment.]
In which case, Austin’s line of reasoning is correct.
If you wanted to do an analysis around the CG (one of those things you could do “just for fun”), you’d need to account for friction on both legs. As you’re working with floor level as a pivot point, neither leg’s friction has any effect–except to help generate a moment, which for this particular problem is negligible (PVC on carpet not having a very high coefficient of friction).
Thanks again folks. The “lesson” ran without a hitch - I was surprised how effective it was! The goal of the exercise was to enumerate the design parameters that they could optimize, and in that respect it was perfect.
If you folks want to re-run it back at home, here is the 30 second overview of the 10 minute exercise:
Super brief intro to torque. We had a mix of students, many of whom had not taken any physics yet. Just enough meat to convince them that torque = force * distance.
Draw the box on the whiteboard, show the forces. Label everything that matters, nothing that doesn’t.
Convert to torques.
Convince them that the box tips when the total torque is in the tipping direction.
Convert the torques to an inequality.
List the design parameters they can tweak at this point. Our first iteration was:
height of pusher
distance from center of gravity to pivot point
Break apart the second design parameter.
How can you move the center of gravity?
How can you move the pivot point?
((There was a 1 minute unscheduled brainstorm that I let run a bit, and then pulled back in. With 15 students, I didn’t want them to move too far yet. I did not participate in the brainstorm.))
Brief reminder of friction, and understand that in steady state and moving, the pusher force is equal to the friction force. If it gets rammed, it exceeds friction.
Convince students that the formula that was Fd1 < Wd2 becomes Wud1 < W*d2. Cancel W.
From the new equation, list all design parameters:
height of pusher
center of gravity
pivot position
coefficient of friction
((Drivers ramming)) <- not really a design parameter, ish.
At this point I turned the meeting over to the student lead and stepped back. It was really neat to see how quickly they fixed their design, as a team. I’m sure a mechanical mentor could do a much better job, but this suboptimal way was effective and could easily be taught by any vaguely technical mentor.
