Torque Application: Simple Q&A Diagram

[vela]

that confuses me because ... we started with a single external force of 10N and we end up with a resultant linear force of 12.974N and some torque

The object always has the two forces acting on it. When you analyze the rotational motion, you have to choose an axis to calculate the moments about. By choosing the pivot as the location of that axis, the moment due to F_x is 0 regardless of what F_x is. That's exactly why you choose that axis. Doing so eliminates one unknown from the equation and makes it solvable.

 

[kuruman]

understood - I was doing a check to match the output of my simulator
can I take it that my workings are correct?

They are correct up to step 4 which you still need to clarify to yourself.

that confuses me because ... we started with a single external force of 10N
and we end up with a resultant linear force of 12.974N and some torque

No. We started with a linear force F = 10 N and a horizontal force at the pivot (yet to be determined) such that the pivot remains at rest while the rest of the rectangle rotates about it instantaneously.

 

[WildEnergy]

They are correct up to step 4 which you still need to clarify to yourself.

No. We started with a linear force F = 10 N and a horizontal force at the pivot (yet to be determined) such that the pivot remains at rest while the rest of the rectangle rotates about it instantaneously.

ok I think I am beginning to see

I was confused before because the unknown force at the pivot was a reaction to the external force
but because the action was exerted on the pivot - and the reaction was exerted on the rod
it becomes part of the total force on the rod

and I was confused about the relationship between translation forces and rotation forces
for instance in this case a torque has resulted in a reaction that has resulted in a change
of both resultant torque and resultant force (change from what would happen without the contact)

but when calculating the acceleration of a non-slipping circle down a slope the force is shared between the rotation and translation so that a = \alpha r

and the contact force must be 2.974N

 

[kuruman]

Can you finish now? Keep us posted.

 

[WildEnergy]

Can you finish now? Keep us posted.

Thanks for all your help - my understanding is improving :-)

As for finishing - this is just the start - I am writing a vehicle simulator
but both my maths and physics are weak - luckily I do at least know how to write software

I bought a Classical Mechanics book which I am working through now

and this question was related to calculating the lateral friction at the rear wheels based on the steering torque generated by the angled front wheels - that is why I was trying to avoid the pivot concept

I am still keen to find an alternative solution that avoids the parallel axis theorem etc.

 

[yuiop]

Thanks for all your help - my understanding is improving :-)

As for finishing - this is just the start - I am writing a vehicle simulator
but both my maths and physics are weak - luckily I do at least know how to write software

I bought a Classical Mechanics book which I am working through now

and this question was related to calculating the lateral friction at the rear wheels based on the steering torque generated by the angled front wheels - that is why I was trying to avoid the pivot concept

I am still keen to find an alternative solution that avoids the parallel axis theorem etc.

If there is almost no friction such as when the car is on ice and the car is skidding, then it will turn around its COM as if that is the fulcrum. If there is normal friction between the road and the car and no skidding, then the center of the rear wheels axle, acts as the fulcrum. The back wheels effectively act as the bearings for the fulcrum or hinge and there is no lateral friction on the rear wheels due to the turning torque of the front wheels. The way to calculate the lateral friction of the rear wheels is to work out the "centrifugal force" of the turning car and the lateral friction of the rear wheels will be equal to the centripetal force required to keep the car moving in a circle and the pivot concept is irrelevant.

 

[WildEnergy]

Brilliant - that way of looking at it really helps :-) Thanks

 

[WildEnergy]

I found a very easy and intuitive way to solve this by considering the constant relation:
Atotal = Aforce + Atorque
Aforce = F / m
Atorque = F * d * d / I
now it is easy to work out the force at the contact point