If a car is moving with constant speed, the wheels of the car exerts a constant force on the floor. Since $F = ma$, the car should be accelerating rather than maintaining the same speed. What is going on here?
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$\begingroup$ What is your question? Your car output is not constant first of all, and the car does not exert a constant force on the floor. Unless we are talking about a car already accelerated up to a speed, and the engine maintain that speed on the road. Are you aware of frictional forces? $\endgroup$– nammerkageCommented Apr 19 at 7:05
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2$\begingroup$ This may be your answer: physics.stackexchange.com/q/709480 $\endgroup$– Sancol.Commented Apr 19 at 7:27
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$\begingroup$ Toning this down to highest approximations, When you stop putting your leg on Accelerator(Which you say stable engine), Friction on tires reduces the speed of car, to maintain constant speed despite being the friction working, One must use te accelerator,Also, I still have not considered air resistance $\endgroup$– Dheeraj GujrathiCommented Apr 19 at 7:47
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3$\begingroup$ A question which presupposes a falsehood is hard to answer. Cars, like all other objects, do accelerate proportionally to the net force applied. Why do you think they do not? $\endgroup$– Eric LippertCommented Apr 19 at 8:21
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1$\begingroup$ Something I find important and is often confusing. If using $F=ma$, $F$ is always the combined force that comes from summing all forces together. Sometimes I write it $\Sigma F=ma$ to remind myself. If I feel like F=ma isn't working, its a strong indication that I'm failing to account for one of the forces that needs to be summed in. $\endgroup$– Cort AmmonCommented Apr 19 at 17:46
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4 Answers
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The car experiences not only the force made by the engine. There are also other forces: friction and air resistance. The air resistance force is proportional to $v^2$, hence it becomes more and more important at higher speed $v$. So the total force becomes smaller while the car accelerates, until (at a certain high speed) the forces cancel to zero and there is no more acceleration.
answered Apr 19 at 7:31
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The motive force produced by the engine is not constant because of limitations in the mechanical design of engines. Each time a car goes up a gear, it trades some of the motive force for allowable speed such that the product of force and speed (power) is maintained under the mechanical limits of the engine.
The higher the speed of the car, the longer the gear ratio needed to maintain the engine speed its limits. The longer the gear ratio is, the less torque multiplication happens through the drive train and less torque arrives at the wheels.
After torque is converted into motive force by the tires and the contact patch, you must also include other forces that apply to the car to get to use $F = m a$. Other forces include some portion of the weight of the car if the car is moving up a grade, aerodynamic resistance when the car is moving at higher speeds, rolling resistance of the tires, friction, and engine compression as well as other mechanical losses through the drivetrain.
Broadly speaking, considering the motive force of the engine and the air drag is sufficient to apply Newton's law of motion
$$ F_{\rm engine} - F_{\rm drag} = m_{\rm car} a_{\rm car} $$
and if the car is going up a grade of $\theta$ degrees
$$ F_{\rm engine} - F_{\rm drag} = m_{\rm car} ( a_{\rm car} + g \sin \theta) $$
Very broadly you can estimate the motive force of the engine from the power it produces at any moment
$$ F_{\rm engine} = \frac{P_{\rm engine}}{v_{\rm car}} $$
and the aerodynamic force that increases parabolically with speed
$$ F_{\rm drag} = \beta \; v_{\rm car}^2 $$
answered Apr 19 at 8:26
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$\begingroup$ These formulas should include somehow the influence of combustion (mass changes, after all)... $\endgroup$– Radek DCommented Apr 26 at 12:57
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$\begingroup$ @RadekD - For a vehicle this has a secondary or tertiary effect at most. A more detailed aerodynamic model and accounting for driveline losses is orders of magnitude more important that the few grams of exhausted burned fuel that is released. $\endgroup$ Commented Apr 26 at 13:15
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The car is exerting a force on the road, but there are opposing forces due air friction and road friction and the net force is zero, so there is zero acceleration.
While f = ma implies acceleration should always be associated with acceleration it is easy to see this is not always the case for individual forces (rather than net forces) when you consider applying a force to a wall and there is no acceleration or movement. There is an equal and opposite reaction force from the wall and the net force is zero.
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If a car is moving with constant speed, the wheels of the car exerts a constant force on the floor. Since F=ma , the car should be accelerating rather than maintaining the same speed. What is going on here?
A car moving in the x (horizontal) direction at constant speed means that $a_x = 0$ so there is no net force in the x direction and no change in velocity.
If the car's motion is pure rolling (no slipping), then the frictional forces between the tire and the road dont contribute to the motion in the x direction.
In the y (vertical) direction, gravity is acting downward on the car (through the wheels) but it is countered by the ground which exerts a Force that is equal and opposite to the gravitational force so there is no net force in the y direction and no acceleration or change in velocity.
The sum of all the forces is zero so the car maintains a constant velocity (ignoring drag and internal losses like gear friction).
answered Apr 19 at 17:42