I need some help in designing my robotic car. So its going to have 4 wheels, each driven by a 12-volt motor. It occurs to me that the weight of the chassis itself will exert some pressure on the wheels and squeeze them against the ground, right? Thereby increasing friction. So is there a way to calculate the pressure on the wheels, that way I can design the body accordingly.
[Physics] How to calculate pressure exerted on the wheels of a robotic car
classical-mechanicskinematics
Related Solutions
Great question. I'm surprised that upon searching, I haven't come across a train-push vs pull question in Physics SE. I'll try to give a detailed answer.
TLDR; Conceptually, the pulling engine is better but both push and pull trains are doable and exist in real life. If you're talking about an idealised thought experiment, I don't think there's a difference.
Now, let's talk details. Your force reasonings are accurate in that, you do no more work pulling a weight up a hill than you do pushing it. The normal/reaction force that is relevant to the friction experienced is perpendicular to the push/pull force and as such, cannot contribute to friction's magnitude. However, it doesn't really matter since train wheels almost never slip. Of course, there is the occasional slip where ice, grease, organic matter, etc. are concerned but steel on steel with heavy weight makes for some impressive tractive force. See this question for more details.
In real life, many train companies use both push and pull methods. In a push-pull train, you have dual locomotives in the front and back, sometimes working together, sometimes taking turns. Companies also do pull trains one way and then push them back, saving cost and having to turn the train around. If we're talking pull-only vs push-only trains, it's a different story.
Theoretically, no, the engine pushing at the rear will not have any mechanical advantage over the engine pulling from the front. In fact, it is the other way around. For many reasons, I think the engine pulling has the advantage, however small.
Easier to see
For one, it is easier and safer to get where you're going when you place your sensors in such a way that the information relevant to your motion reaches you soonest. Almost always, this is the part furthest along the direction of your motion (which is why most animals have their eyes in front). In other words, you get to see what's in front of you before you hit it.
Easier to make
Secondly, a train that has a pulling engine is much easier to design and build. Most train cars are connected by a "tether" of sorts, which is much closer to a string than it is to a stick. It is a lot easier to design and build connections that are strings that pull cars than sticks that push cars. What I mean by stick is something which is rigid and resists deformation. What I mean by string, however, is something which pulls and is pliable. An interesting aside, I've heard my professor once say that that's sort of the definition of a string in physics; something which can only pull and not push.
Anyway, in real life the train cars don't do well when they are pushed with a string (even a semi-rigid connection). You get collapses and distortions in the overall chain of the train because the connections can't withstand the force intended to push the train. The tracks help mitigate this to some degree but it creates unwanted stress.
Easier to steer/safer to drive
It only makes sense to push the train if the connections are rigid but then steering the train becomes mechanically harder because the train becomes less flexible as a whole. The chances of derailing are also higher in a push-train than it is in a pull-train although I am aware some experts say that the difference is small enough to be ignored and isn't significant (especially in reference to Glendale 2005 and Oxnard 2015). I think this is because the direction of force is changing sooner with respect to the direction of track change in a pull-train than it is in a push-train. In other words, the pull force changes with the curve and the other cars follow accordingly but the push force remains straight as the cars in front experience the curve in tracks.
More efficient design
You also get inefficiencies when you push a non-rigid train because all the small things in a train distort whenever and however they can. Forces and these things in general tend to always take the path of least resistance. A path that is non-rigid is by definition less resistant than a rigid one and so whenever a non-rigid path exists and is pushed, it will bend and buckle in a way that it was not designed to do. This creates more friction, wear, tear, heat, noise and in general, more things to account for. Below is just one of the ways I could think of that a push-train going uphill could go awry.
Additionally, a pull-engine has an inherent superiority to a push-engine. Try this; slowly push a cup with your finger across the table. Eventually, you will "lose" the cup. It might slide to the side or be pushed aside by your finger or twist to avoid your finger. Now try pulling the same cup with your finger through its handle. You'll never lose the cup. Not sure how significant this is when there are tracks but I imagine there's certainly a difference.
Idealistically in a thought experiment, I think there is no difference. You'd need some kind of exotic material though, along with perfect rigidity, perfect trains with perfect connections, flawless tracks, etc.
Edit
In response to the updated question, with a rigid coupling, both the pull and push engines have things resisting the torque "lift" of the train (weight of front load in push-engine and the back portion pushing into the ground in pull-engine). Note that whether the locomotive is front-wheel, rear-wheel or all-wheel drive is relevant. That being said, I still maintain that the pull-engine is superior because the point here is essentially that the train is doing a power wheelie. The best way to mitigate wheelies isn't by moving more weight to the front of the vehicle, it's by adding a wheelie bar.
[...] and there's a large normal force on the system I've described, and it points forward, along the direction that would push the car forward
The normal force points vertically upwards (it comes into existence to balance out the weight). The force pulling sideways is not a normal force but static friction. You say it right later on, but messes up the words here.
But this scenario is only applying to the section of the wheel that is on the asphalt. So, as the wheel turns, different sections of the wheel have this effect. How does this produce a thrust forward for the car if the car keeps losing that segment of the tire that is producing a friction force with the asphalt?
How do you make a thrust when walking? You do that by pushing off from the ground with one foot, and then the other foot takes over. Each contributes with a portion of the forward thrust.
Now imagine a star. When a spike is touching the ground it pushes off from the ground in the same way. It creates thrust, acceleration, while it pushes. Then the next spike takes over.
Now imagine a star with more spikes. With many more spikes. So many that you can't distinguish them. So many placed so close that it looks like... a round object. A wheel. When this wheel rolls, each spike - which is now only a single point - pushes off from the ground during the infinitely short time it is touching.
It might be a very short time of thrust done by one point on the wheel, but it happens constantly, since another point takes over immediately. In total, there is constantly a thrust going on.
Is there a tangential force fighting against $Ff$ that I haven't included?
No. (In realistic circumstances there will be some energy loss in axles and gearing, in the flexing rubber and in a deforming surface - all such losses will slow down the wheel and are in one term called rolling resistance. But in ideal models these are neglected (just like air drag is), and only static friction is working horizontally.)
Because the motor is accelerating the tires radially, [...]
A clarification on the wording: The motor is accelerating the tires by applying a torque around the axle. This makes the tire turn. Because all the particles of the tire are stuck together, they hold on to each other inwards - radially. This is why the radial acceleration appears on them. Yes, a particle in circular motion must have a radial acceleration, but the motor is only indirectly causing that - the direct reason is rather the rigid wheel and the hold-on forces from inner particles.
[...] so each segment of the tire should have a tangential force on it that is going to be, at the moment I've described, opposite to friction?
And to the question: Yes, at the contact point where static friction happens there is a force oppositely. The wheel is applying a static friction on the ground. The ground replies with a static friction on the wheel in the other direction. This is Newton's 3rd law.
But there are no other forces on the particles of the wheel than this friction.
There are of course still the holding-on forces between particles as mentioned above, which pull every particle towards the centre - but this makes no difference in the motion, since at the contact point these are perpendicular to the friction force and direction of motion and have no influence.
In the same way there are of course also still weight and a normal force, but these also work at (or through) the contact point perpendicularly to the motion and have no influence.
What if there was no motor force on the wheels, and it just had some $\omega$? This scenario I've described in the latter sentence sounds like asking if there is a friction force acting on an object moving at a constant speed.
By "no motor force" you mean no motor torque. No motor torque means no forced push. The wheel touches the ground but does not try to push itself off from the ground, because it doesn't try to speed up the rotation. When there is no pushing off from the ground, there is no need for friction to appear to hold back against anything.
Conclusion: There is no static friction happening when the wheel rolls without accelerating. In an ideal case, it will roll with constant $\omega$ forever.
Best Answer
You have to know where the center of gravity is. If $a$ is the % distance along the wheelbase for the center of gravity (50% = center, 0% = front, 100%=back), and $b$ the % distance along the track for the center of gravity (50% = center, 0% = left, 100% right) then the weight fractions for each wheel are:
$$ (\mbox{front-left}) = (\mbox{Weight}) \frac{3-2a-2b}{4} $$ $$ (\mbox{front-right}) = (\mbox{Weight}) \frac{1-2a+2b}{4} $$ $$ (\mbox{rear-left}) = (\mbox{Weight}) \frac{1+2a-2b}{4} $$ $$ (\mbox{rear-right}) = (\mbox{Weight}) \frac{2a+2b-1}{4} $$
There equations come from the balance of moments in two planes. Here is a top view of the balance.
Example:
I the front back-balance is $a=0.4$ and the left-right balance is $b=0.55$, with a chassis weight of $0.5\,{\rm kg}$ then the corner weights are:
$$ (\mbox{front-left}) = (0.5) \frac{3-2*0.4-2*0.55}{4} = 0.1365 {\rm kg}$$ $$ (\mbox{front-right}) = (0.5) \frac{1-2*0.4+2*0.55}{4} = 0.1625 {\rm kg}$$ $$ (\mbox{rear-left}) = (0.5) \frac{1+2*0.4-2*0.55}{4} =0.0865 {\rm kg}$$ $$ (\mbox{rear-right}) = (0.5) \frac{2*0.4+2*0.55-1}{4} =0.1125 {\rm kg}$$
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