What Is the Main Force Driving a Car's Acceleration?

  • Thread starter Thread starter davon806
  • Start date Start date
  • Tags Tags
    Accelerating Car
AI Thread Summary
The main force driving a car's acceleration is static friction, which acts on the driving tires to propel the vehicle forward. When the wheels rotate backward, they exert a backward force on the ground, and according to Newton's third law, an equal and opposite force from the ground accelerates the car. Some textbooks illustrate friction with a backward arrow to represent opposing forces like air resistance and rolling friction, which slow the car down. However, the friction force that contributes to acceleration is the traction force acting on the driving tires. Understanding these dynamics clarifies the role of friction in a car's movement.
davon806
Messages
147
Reaction score
1

Homework Statement


When the wheels rotate backwards,a backward force is exerted on the ground by the tyres.
By 3rd law,a equal but opposite force acts on the tyre by the ground,so the car
accelerates.
But I have also found some websites that states static friction is the only external
force causing the car to move forwards.
In some typical physics questions,the driving force and friction are two different
forces act on a car,where friction is in the opposite direction of displacement.
I would like to ask which force is actually accelerating the car?Friction or reaction
force by the ground(due to the engine)?
If it is friction,why backward arrow is drawn in some textbooks to represent friction while
the car is accelerating forwards?


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
davon806 said:

Homework Statement


When the wheels rotate backwards,a backward force is exerted on the ground by the tyres.
yes, this is a friction force
By 3rd law,a equal but opposite force acts on the tyre by the ground,so the car
accelerates.
yes, by friction
But I have also found some websites that states static friction is the only external
force causing the car to move forwards.
that is correct
In some typical physics questions,the driving force and friction are two different
forces act on a car,where friction is in the opposite direction of displacement.
the driving force (or traction force) is the friction force on the driving tyres acting forward; friction on the non driving tyres acts backwards
I would like to ask which force is actually accelerating the car?Friction or reaction
force by the ground(due to the engine)?
friction is the reaction force, accelerating the car
If it is friction,why backward arrow is drawn in some textbooks to represent friction while
the car is accelerating forwards?
maybe they are showing the forces acting on the ground by the tyres and not the force on the tyres from the ground..?
 
davon806 said:
If it is friction,why backward arrow is drawn in some textbooks to represent friction while
the car is accelerating forwards?

That's probably showing the other friction forces like air resistance and rolling friction that are tending to slow the car down and so act backwards.

That's separate from the reactionary friction force on the wheels driving the car forward.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top