- #36

russ_watters

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Hopefully the other mods won't be upset with me, but after several days...could you draw it out for me? I am just confused.

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- Thread starter alkaspeltzar
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- #36

russ_watters

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Hopefully the other mods won't be upset with me, but after several days...could you draw it out for me? I am just confused.

- #37

russ_watters

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From the link:Originally, I wanted to know how the tire got the 'strength' to push backwards. This link makes sense if it is true. Can anyone verify this.

https://www.school-for-champions.com/science/friction_rolling_starting.htm#.XYo7bBhOk0M

Russ Watters, if I am not wrong, this was your point above.

Yes, I did make the analogy in the other thread of a stick falling over. Friction prevents the bottom of the stick from sliding backwards, thus pushing the stick forwards.The force of the static sliding friction prevents the wheel from sliding and thus initiates the rolling motion. The rolling motion is actually a form tilting about the point in contact

- #38

alkaspeltzar

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Russ, so am I seeing this correct? The force from the axle creates torque on the wheel. The friction force creates an opposite torque. There is a net torque, they don't cancel completely, so wheel experiences angular acceleration.

IF we back out and look at the car as a whole(engine, axle, tire etc), it is the friction force which unbalances the force horizontally and propels the car forward.

- #39

alkaspeltzar

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From the link:

Yes, I did make the analogy in the other thread of a stick falling over. Friction prevents the bottom of the stick from sliding backwards, thus pushing the stick forwards.

Thank you Russ. This is very calming to finally understand! Sorry for the banter back and forth all.

- #40

A.T.

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From the tires, obviously.Where does the force from the tires on the road come from?

Your attempt to relate forces and torques using cause-effect-reasoning, is fundamentally misguided and will not help you to correctly analyze mechanical setups. See this little toy for example, and try to build a chain of what is causing what:

- #41

alkaspeltzar

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Part of what keeps getting me confused is I keep going back to Newton's third law and involving those forces. I know we shouldn't but I think well how does this force cause this or that and it get messy real quick

- #42

Dale

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As far as I can tell you never even attempted to apply Newton’s laws, despite my repeated recommendation to do so.Part of what keeps getting me confused is I keep going back to Newton's third law and involving those forces.

That last statement is the answer to the question. The use of free body diagrams is designed to help organize the analysis of a complicated problem by breaking it into smaller manageable systems. You follow the rules because doing so systematically and consistently allows you to answer questions like this reliably and without guesswork.I think one thing that would help is understanding why when doing a free body diagram do we ignore all the internal forces?

...

I know we shouldn't but I think well how does this force cause this or that and it get messy real quick

- #43

alkaspeltzar

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Well I do Newton's laws. I do a lot of design everyday with them. But lately on questions like this I see myself mixing internal and external forces and creating confusion. I've been over analyzing physics and trying to dig myself out of a holeAs far as I can tell you never even attempted to apply Newton’s laws, despite my repeated recommendation to do so.

That last statement is the answer to the question. The use of free body diagrams is designed to help organize the analysis of a complicated problem by breaking it into smaller manageable systems. You follow the rules because doing so systematically and consistently allows you to answer questions like this reliably and without guesswork.

It's been a long time since I took physics, so my diagrams and access to help are limited, which is why I ask for help. So please understand I'm not trying to not listen .

- #44

rcgldr

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The car is pushed forwards, but the acceleration of the car coexists with the angular acceleration of the drive train. As I posted before, a small part of the torque is performing angular acceleration of the drive train, while as you posted, the majority of the torque translates into the N3L pair of forces between road and tires.So some torque is being used to accelerate drive train and wheels. I agree, that doesn't contribute to the N3L reaction/action pair of the road and tires.

So that means the rest/majority of the torque from the engine is being applied as the force of the tires onto the road right? And then because of this, the car is pushed forward by the static friction.

That website is not taking acceleration into account, as none of that torque is shown as being opposed by the angular inertia x angular acceleration.RCGLDR, is this website below correct as well. I think that explains where the majority of the torque is going, applying the force to the road.

https://www.school-for-champions.com/science/friction_rolling_starting.htm#.XYo7bBhOk0M

So the torque input into the wheel = (angular inertia x angular acceleration) + (N3L force x radius).

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- #45

Dale

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The suggestion to write down the equations was help, which although you asked for you did not take advantage of when it was offered. Tell me, how do you believe that every expert on this forum gained their expertise? Do you not recognize that it is by doing exactly what I recommended?I've been over analyzing physics and trying to dig myself out of a hole

It's been a long time since I took physics, so my diagrams and access to help are limited, which is why I ask for help.

You did the free body diagram and showed good effort there. Immediately you got the same valuable feedback from two separate sources, guiding you on how to do this work correctly. As a result you will be able to do this better in the future.

Now, the next step is to use the corrected free body diagrams (post 36) to write down Newton’s laws. You will make mistakes and we can correct those mistakes as you make them. But until you write down your understanding we cannot know where your analysis is going wrong. The purpose of the exercise is so that you can develop understanding that only comes through doing such exercises, making mistakes, and fixing them.

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- #46

russ_watters

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No, the force is just a force. The axle applies a forceRuss, so am I seeing this correct? The force from the axle creates torque on the wheel.

My FBD does not include numbers, so whether or not there is an acceleration depends on the numbers. This FBD works for constant speed and for acceleration.The friction force creates an opposite torque. There is a net torque, they don't cancel completely, so wheel experiences angular acceleration.

Internal forces are internal, so they can't cause an object to move...though it is probably more complete to just say it's a convention used to organize thoughts, as Dale says. The convention is chosen based on what the people who created it want to use the FBD to do.I think one thing that would help is understanding why when doing a free body diagram do we ignore all the internal forces? We only focus on external.

- #47

A.T.

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Yes, so stop it.I think well how does this force cause this or that and it get messy real quick

- #48

Dale

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##F_a-F_w=m_c a##

Newton’s linear 2nd law at the wheel is

##F_f-F_a=m_w a##

Newton’s rotational 2nd law at the wheel gives

##I_w \alpha = T-F_f r##

Assuming no slipping we have the constraint

##r \alpha=a##

Our knowns are ##T##, ##r##, ##m_c##, ##m_w##, and ##I_w##. Our unknowns are ##F_a##, ##F_f##, ##F_w##, ##a##, and ##\alpha##. So we have four equations and five unknowns. We need one more equation, such as the wind force ##F_w##. If we are starting at rest then ##F_w=0## otherwise we could have some equation that gives ##F_w ## as a function of the velocity. For simplicity let’s just assume ##F_w=0##. Then we have four equations in four unknowns, which we can solve. When we do that we get $$a=\frac{r T}{I_w+(m_c +m_w)r^2 }$$

We see that this is never zero unless ##T=0##

- #49

alkaspeltzar

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What force is F sub a? Thanks Dale!

##F_a-F_w=m_c a##

Newton’s linear 2nd law at the wheel is

##F_f-F_a=m_w a##

Newton’s rotational 2nd law at the wheel gives

##I_w \alpha = T-F_f r##

Assuming no slipping we have the constraint

##r \alpha=a##

Our knowns are ##T##, ##r##, ##m_c##, ##m_w##, and ##I_w##. Our unknowns are ##F_a##, ##F_f##, ##F_w##, ##a##, and ##\alpha##. So we have four equations and five unknowns. We need one more equation, such as the wind force ##F_w##. If we are starting at rest then ##F_w=0## otherwise we could have some equation that gives ##F_w ## as a function of the velocity. For simplicity let’s just assume ##F_w=0##. Then we have four equations in four unknowns, which we can solve. When we do that we get $$a=\frac{r T}{I_w+(m_c +m_w)r^2 }$$

We see that this is never zero unless ##T=0##

- #50

russ_watters

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Force on or by the axle, from my FBD.What force is F sub a? Thanks Dale!

- #51

alkaspeltzar

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Okay thanks. I wasn't taking into account there was a force from the axle. I always was thinking that if the torque was close to zero, then the forces would balance too. But I see that isn't the caseForce on or by the axle, from my FBD.

- #52

russ_watters

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If what torque is close to zero? Providing a large torque to the wheels at high rpm - large enough to move a 2000kg car at high speed - is most of what the engine is for!Okay thanks. I wasn't taking into account there was a force from the axle. I always was thinking that if the torque was close to zero, then the forces would balance too. But I see that isn't the case

- #53

Dale

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I only solved for ##a##, but I encourage you to solve for the unknown forces for your own experience.I always was thinking that if the torque was close to zero, then the forces would balance too. But I see that isn't the case

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