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I have 2 related questions, the second one is more engineering than physics, but it relies on the first.

Does it take more force to get a car to start moving than to accelerate it the same amount while it is already moving?

For example, to get it from 0 to 5 mph vs 40 to 45 mph.

I ask this because the gear ratio of car gears is higher for low gears. This does two things, it increases the torque (and wheel force) in lower gears and increases the rpm's of the engine for lower wheel speeds.

Is the only reason high gear ratios are used for lower gears to raise the engine rpm's from the wheel rpm's (because engine speed wants to be around 2-4000), or is it because you need that additional torque to get the car to start moving.

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  • $\begingroup$ There are too many questions here for a single post. Pick the one you are most strongly interested in. Then if you still have another question, post that. And when you start talking about automotive torque and gear ratios, that borders on engineering, so don't be surprised if your question get's migrated or closed. $\endgroup$ – Bill N Apr 25 at 22:25
  • $\begingroup$ Maybe you should look up Newton's Second Law? $\endgroup$ – Gert Apr 25 at 22:25
  • $\begingroup$ @Bill N yeah I just realized that. I will edit it $\endgroup$ – Splry00 Apr 25 at 22:28
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    $\begingroup$ @Gert I don't think it is simply F = ma here because the opposing forces (friction) may change as velocity increases. That's why I ask. $\endgroup$ – Splry00 Apr 25 at 22:33
  • $\begingroup$ It takes more energy to accelerate from 40 to 45 mph then from 0 to 5 mph because the air friction is greater in the 40-45 mph region. $\endgroup$ – Shamaz Apr 25 at 22:59
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The car engine must run at a minimum of approximately 750 rpm to remain running. This means that you can't easily get the car moving in high gear because the low starting torque in that gear would stall the engine. If you want to test this out, learn how to drive a standard transmission, and from rest, release the clutch when the transmission is in the highest gear. You will find that the engine stalls when you do this.

Regarding the force involved, this question can't be answered in a simple fashion. According to Newton's 2nd law, $a=F/m$, so a smaller force will give a smaller acceleration. It's more meaningful to talk about the amount of work involved to change the car's velocity by a given amount. The work/kinetic-energy theorem states that the amount of work involved is equal to the change in kinetic energy, assuming no dissipative forces (e.g., friction) are involved. Under this assumption, it is obvious that there is a larger change in kinetic energy, and hence more work involved, in going from 40 mph to 45 mph than there is in going from rest to 5 mph.

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Does it take more force to get a car to start moving than to accelerate it the same amount while it is already moving?

No. $F=ma$ holds at all speeds. What you are noticing is that it is increasingly difficult (more power is required) to create such a force as the relative speed increases.

Imagine you are standing next to a merry-go-round. While it is stopped, you have no difficulty applying a reasonable force to it, which causes it to accelerate. Once it starts moving quickly, you will be unable to apply the same force. The acceleration slows because the forces are not constant.

Is the only reason high gear ratios are used for lower gears to raise the engine rpm's from the wheel rpm's (because engine speed wants to be around 2-4000), or is it because you need that additional torque to get the car to start moving.

It's mostly to keep the internal combustion engine's rotation speed in a useful range.

Compare instead an electric motor. It won't have the speed limitations of a combustion engine. It can deliver full torque when $v=0$. But (assuming constant power) as the speed of the motor increases, the torque that it can deliver decreases. Or said another way, it requires increasing power to deliver constant torque.

https://www.engineeringtoolbox.com/electrical-motors-hp-torque-rpm-d_1503.html

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