Magnitude of friction on the driving wheel

[sgstudent]

After discussing this with my friend using our O level knowledge, we came to a conclusion. But we have some queries.

So firstly, we have our driving wheel on the air and the engine is turned on. So the Free Body Diagram would look like this: http://imgur.com/xVk0w

So in this case the net force of the wheel is 0N as F2 and F1 are equal in magnitude. However, about F1, there is a net moment which turns clockwise (F1XL).

Then now we put it on the ground and and experiences a friction oppsoing F1 as friction opposes force in a single plane of contact. So now, the Free Body Diagram would look like this: http://imgur.com/Pw1oh

From this, since F2-F1=0N, so the net force=friction only. While the wheel still has to maintain its clockwise moment so F1L > FrictionL and since L is a constant, so F1>Friction.
So from here, it all makes sense to us, but then after thinking again since that friction force that we have been dealing up to now is static friction so shouldn't F1 be equal to friction? But if F1 is equal to the friction, the the wheel would not be able to turn in the first place.

So we were stuck at this part with no solution in mind. Could someone help us out with this? Thanks :-)

 

[eljose79]

I would like to provide some clarification and further explanation on the concepts discussed in the forum post.

Firstly, the Free Body Diagram is a visual representation of all the forces acting on an object. In the case of the driving wheel on the air, the net force is indeed 0N as F1 and F2 are equal and opposite. However, it is important to note that this only applies in a frictionless environment. In reality, there will always be some form of friction present, even if it is minimal.

When the driving wheel is placed on the ground, it experiences a frictional force opposing F1. This frictional force is necessary for the wheel to turn, as it provides the necessary torque to overcome the moment created by F1. Without this frictional force, the wheel would not be able to turn.

It is correct to say that in this scenario, F1 is greater than the frictional force. This is because the wheel needs to maintain its clockwise moment, and in order to do so, F1 must be greater than the frictional force.

It is important to note that friction is a complex force and can be affected by various factors such as the surface material, weight of the object, and the applied force. In this case, the frictional force is likely to be a combination of both static and kinetic friction, as the wheel is both rolling and sliding on the ground.

In summary, while it may seem contradictory that F1 is greater than the frictional force, it is necessary for the wheel to maintain its motion and overcome the frictional force. Friction is a crucial force in many everyday scenarios, and understanding its effects is essential in the study of mechanics. I hope this helps clarify any confusion and feel free to ask any further questions.