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We say acceleration is produced in a body when a force acts on it in some direction.When a car moves with acceleration and I'm an observer on ground.What force should I say is acting on the car?

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  • $\begingroup$ Note that in Newton's 2nd Law of motion, $\mathbf{F} = m \mathbf{a}$, the symbol $\mathbf{F}$ is the NET force acting on the body with mass $m$. $\endgroup$ – K_inverse Mar 22 at 7:39
  • $\begingroup$ The wheels(rear) exert a force against the ground in a horizontal direction. For a schematic please check out this link <auto.howstuffworks.com/auto-parts/towing/towing-capacity/…> $\endgroup$ – Nixnyete Mar 22 at 7:53
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The engine applies torque to tires and they apply forward directed force the car's axel by friction with road.

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  • $\begingroup$ The tyre force is not a friction force, it a function of the slip condition $\endgroup$ – Eli Mar 22 at 13:26
  • $\begingroup$ @Eli, I'm not sure your comment clears anything up. Maybe you could expand it into an answer? $\endgroup$ – BowlOfRed Mar 22 at 15:56
  • $\begingroup$ @BowlOfRed this is the equation for the tyre force $F_s$, where s is the sleep I don’t see any relation to the friction? $\begin{aligned}F_{s}=f\left( s\right) \\ s=\dfrac {v-w\cdot r}{\left\| v\right\| }\end{aligned}$ $\endgroup$ – Eli Mar 22 at 18:15
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The road is applying a force of friction on the tires at the line* of contact. This force is carried to the axles and to the chassis due to their rigidity (approximately).

It might be a bit confusing to use active language for the road, but the road's application of force is simply a result of the "action-reaction" principle. The tires want to speed up due to their connection to the engine. But this applies a force to the road. So the road applies an equal but opposite force to the tires.

*If the contact between the tires and the road is a line (ideal), then you have static friction, since there is no relative motion between the road and the tires at the line of contact. However, of course the contact is more a patch of several square inch, and most of the tire is rubbing against the road (i.e., relative motion) so you have mostly kinetic friction.

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