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The car is a 2002 Toyota Corolla SE I have in my garage on a stone surface, I.e., the floor itself.

What I want to know is how to determine the force needed to get it in motion at all.

The curb weight is 2,400 lbs. How could I determine the force needed to push it on neutral to get it in any kind of motion?

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closed as off-topic by Chris White, Dan, Dilaton, David Z Jul 13 '13 at 8:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

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The biggest factor is probably all the internal friction in the vehicle (without friction it would require no force at all), and as such not answerable by us. –  Chris White Jul 13 '13 at 3:06
    
I don't understand what friction you aelre referring to –  legate stroke Jul 13 '13 at 3:07
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@ChrisWhite but you don't think it should be closed? –  David Z Jul 13 '13 at 3:16
    
@DavidZaslavsky well, I figured maybe there was a chance there was an easy non-engineering solution ::shrug:: legate stroke: any surfaces in contact and moving relative to one another lose energy to friction. All the moving parts in the car contribute, and without this effect, you could push a car with your little finger if you so chose. –  Chris White Jul 13 '13 at 3:20
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@Chris I dunno, I think if you have to italicize "maybe" then that's not really enough to justify leaving it open ;-) I mean, if you really think the question is a good one and appropriate to remain open, by all means, don't VTC, but it seems to me like this falls under the engineering category, or insufficient prior research. –  David Z Jul 13 '13 at 3:25

2 Answers 2

You may have come across the terms static friction and dynamic friction. In brief, the force needed to get an object moving is generally less that the force needed to keep it moving. Cars show this phenomenon, though for different reasons to the usual lab experiments of sliding blocks around. Car bearings are designed to maintain a thin film of oil when they're moving, but when the car is stationary this film collapses and the friction rises considerably. The point of this is that Maxim's idea wouldn't give you a good idea of the force needed to get the car going, though it would give you the dynamic friction.

This is one way of doing it:

car

Drive the car onto a platform of length $d$, then release the handbrake and put it into neutral. Now start jacking up one end of the board, and measure the height $h$ at which the car just starts to roll. (You might want someone in the car to stop it! :-)

If the mass of the car is $m$, then the force propelling the car forward is $F = mg\sin\theta$, where $\theta$ is the angle the board makes to the floor. The value of $\sin\theta$ is given by:

$$ \sin\theta = \frac{h}{\sqrt{h^2 + d^2}} \approx \frac{h}{d}$$

where the approximation is good if $h \ll d$.

So the force required to start the car moving is:

$$ F = mg \left( \frac{h}{\sqrt{h^2 + d^2}} \right) $$

By just measuring the distances $h$ and $d$ you can calculate what force you'll need to start the car moving.

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You seem to have not finished your point or explanation. –  legate stroke Jul 13 '13 at 18:56
    
@legatestroke: I've edited my reply to give the final formula for calculating the force –  John Rennie Jul 14 '13 at 6:52
    
Giving me that formula is very pointless; I have no idea on how to use it. –  legate stroke Jul 15 '13 at 3:10

We can estimate as follows: Consider the case when the car motor is engaged which results in the car moving at a slow speed. The power delivered by the motor P is the product of the motion speed V times the cumulative resistance force F (including internal friction etc). This force F is exactly what we are looking for, that's the force we need to apply if we push the car with our hands at this speed. So, F=P/V, and let's put the numbers (by the order of magnitude): P~1 hp~1e3 W; V~1 m/s. Then the force is F~1e3 N which is equivalent to ~100 kg weight. This makes sense, two average guys pushing with the force corresponding to 50-75% of their weight should be able to push the car at ~1 m/s.

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Not a fairly believable formulated result. I can not even bench press my bodyweight and I managed to push the car I mentioned in the body with two people in both front seats through grass for ten feet. According to your formula, and adding 450 lbs for the two people, I would need to apply more than 240 lbs of forward pushing force. Not plausible considering I can not even bench press 150 lbs. –  legate stroke Jul 13 '13 at 18:53
    
And I got the car moving faster than 1 M/PH. –  legate stroke Jul 13 '13 at 18:54
    
My calculation uses as an input the value of engine power for moving at a particular speed. I took 1 kW as an order of magnitude for the mechanical power needed for very slow motion of the car; but it could be 0.5 or 5 kW perhaps. Note that bench press is done with the arms only while pushing the car is done with the legs, much bigger muscles involved. For example, you probably can do a sit-up which means your legs can push your weight without problem. What is M/PH - mile per hour? I was talking about m/s - meters per second. –  Maxim Umansky Jul 15 '13 at 2:27
    
Sit ups have no bearing on leg strength totality; the abdominals and obliques are used mostly. Also, you are saying leg strength results in this? You are stating two men can push the car faster tha me? Is that an insult to my strength? I am a powerlifter/strength trainer. –  legate stroke Jul 15 '13 at 3:15
    
Check out powerlifting.stackexchange.com ;) –  Maxim Umansky Jul 15 '13 at 4:33

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