Calcing Rounded Objects

[Flashlight237]

Something I've always been confused about is the ability to calculate round objects, like wheels (and cylinders in general) and, god forbid, spheres. Like, we know about the idea of using a friction coefficient when pushing a non-rounded object like cubes, pyramids, and whatnot, but I couldn't for the life of me remember a single time pushing anything rounded was calc'd. I know that less force would be required because wheels were invented to make moving things easier, but how exactly does one calculate pushing something like, say, a 10-tonne wheel or a 10-tonne sphere?

 

[DarkDragonMedeus]

We do take the friction of the object into account before we calculate force required to move it; which an object's shape can effect that. But after that, we use the F = MA formula.

 

[Jasonsith]

Rolling an object is easier than sliding it (and easier than lifting it up) due to friction to overcome before the object even moves on the ground - a 200 kg cannon on ancient wheels needs 3 soldiers to lift up but only one soldier to push.

Then, F = (M A) + (M G Ffrict) applies.

Note rolling friction is usually lower than sliding friction.

 

[Ari64-SP]

Rolling an object is easier than sliding it (and easier than lifting it up) due to friction to overcome before the object even moves on the ground - a 200 kg cannon on ancient wheels needs 3 soldiers to lift up but only one soldier to push.

Then, F = (M A) + (M G Ffrict) applies.

Note rolling friction is usually lower than sliding friction.

Wait, but isn’t the formula for rolling objects F = Crr*Nf ? And isn’t Crr just a/g? So wouldn’t that make the equation F = (a/9.81)*(m*9.81)?

 

[Jasonsith]

Wait, but isn’t the formula for rolling objects F = Crr*Nf ? And isn’t Crr just a/g? So wouldn’t that make the equation F = (a/9.81)*(m*9.81)?

There is a force in overcoming rolling friction and a force in accelerating the object in general. Just if you do assume negligible force in acceleration then the force to overcome friction remains.