# Confusion with the example of Newton’s first law

1. Jul 8, 2009

### steevan78

Consider a case where 2 people are in a car driving in a straight line at sufficient speed. At some point the driver takes an abrupt 90 degree right turn and owing to reasons attributed to Newton’s first law the co-passenger swerves into the driver (i.e. to the left).

My question is why doesn’t the driver feel as much swerve as felt by the co-passengers to the left (towards his door)? Is that because he is firmly holding the wheel? Or is it that I am not a keen observer?

2. Jul 8, 2009

### fatra2

Most probably that the driver was ready for the abrupt turn, for which he could hold on very tight to the wheel. Otherwise, he would be pushed against the door.

The same would have happened to the passenger. If the passenger could get ready by holding to the holly crap bar, the virtual force could be counter act.

Cheers

3. Jul 8, 2009

### kevinfr0st

the driver does he is just fighting it you still feel the force of you wanting to go on in a straight line the fule in tank would be doing the same as well

4. Jul 8, 2009

### dE_logics

Why do you thing he does not?

5. Jul 8, 2009

### vin300

No, the driver on the left must swerve more during a right turn because at this point, the right of rear and front tyres have lesser velocity than the left, so by F=mv^2/r, he experiences greater centrifugal force.
Or maybe both experience the same force as r is also lesser for the right tyres.
You cannot turn your car abruptly by 90 degrees, can you?
The angular velocity of all the tyres during a curve is the same, so by equation
F=mr*(omega)^2, the driver as wall as the one sitting behind him to the left experience more force.

Last edited: Jul 8, 2009
6. Jul 9, 2009

### kevinfr0st

i think ur looking at it all wrong way (im just a fool who cant spell so take what i say with a pinch of salt) but tyres are nothing to do with it all,to be honest the car has nothing to do with it,if u made the idear simple as you can for simple people like me :) if it was a trolly with 2 tins of beans and you did the experiment my simple head says the 2 tins will carry on forward at equal force
but like i say what do i know

7. Jul 9, 2009

### vin300

When the trolley curves, it goes on changing direction at every point on the curve.The trolley has the same speed, but different velocity(velocity is a vector)at every point on the curve.
To produce a constant change in velocity there must be a constant acceleration(constant force) which if you want to understand, by newton's thid law is exactly equal and opposite to the force experienced by the person in the car causing swerving.
Where does this opposite force come from?
It is the frictional force between the tyres of the vehicle and the road, which is always equal in magnitude to the force a person in the car experiences. It is directed to the centre of the curvature so you are pushed radially outward.
Next, you can read the last two lines of my previous reply.

8. Jul 9, 2009

### kevinfr0st

i still think ur adding more in to it then ur supposed to the question, isnt possable,it is a example,the car and 2 passengers were used to get it in to visual contex,if there was a speed put in to it all ect then it would be about the car ie if to fast the car will spin,even with the trollys and tins of beans it dont totaly give true example because the weight of beans and friction contact with trolly may pull the beans off corse a bit and might even spin the tins,lol was just thinking of the old war machens that had A frame and counter weight on long rotating arm the arm and weight only goes in a round motion but the object gets thrown forward isnt that the same but in reverse,
this is only my thoughts on the matter and take it with a pinch of salt

9. Jul 9, 2009

### vin300

What you have written is difficult to read and understand.
I think that I have mentioned everything i should've.
Yes the tins may rotate outwards on the top and the reason for this is that friction pulls it inside at the bottom and the same centrifugal force results in highest torque on the top(outward) and the tin is thrown away, but this happens only when its velocity is greater than the value sqrrt{(mu)*r*g} where mu is the coefficient of friction between the tin and the trolley, r is its distance from the centre of curvature, so the tin on the left during a right turn is thrown away more easily as it experiences greater force.

Last edited: Jul 9, 2009