A car weighing 13,000 N is driving at a speed of 50 km/h, overcoming a resistance (Fd) of 450 as shown in the figure. If the coefficient of friction between a wheel and a road on a frozen road is 0.065, explain which of the front and rear wheels is good for maintaining speed using the force of reaction and friction.
How should I explain this question?
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1I'm guessing whoever set the question is looking only for a computation involving balancing moments and vertical forces, then noting which wheel can sustain the greater horizontal traction, not for the integrated-engineering answer "if the coefficient of friction is that low, you should have stayed at home". – Daniel Hatton Dec 19 '21 at 11:54
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... and in any case, is "coefficient of friction" meaningful here? ISTR rubber-ice contacts follow Archard's law, not Amontons' law. – Daniel Hatton Dec 19 '21 at 11:55
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Yes. Explain the question in more detail and [edit] it it show what you have tried. – hazzey Jan 09 '22 at 16:18
2 Answers
I think the point of the question is to find out if the car should be front or rear wheel drive (or both or AWD) in order to be able to maintain the speed.
It seems like the question assumes a steady state driving condition which will result in a weight distribution on the front and on the rear wheels (i.e. different reaction on the front $N_F$ and the rear $N_R$ wheels).
Those reaction forces in combination with the friction coefficient will be able to produce only a certain amount of friction (before the wheels start to skid).
That is all I can say without actually solving the question for you.
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The weight is distributed to the rear tires by taking moment about the front axel is:
$$M=13000N *1250mm/3000mm =5410N \ therear \ load, \\ front \ load \ 13000-5410=7590N$$
the drag force is distributed like a couple:
$$Fd_{reaction}= 450N*750mm/3000mm=112.5N \\ \text{112.5N must be added to rear tires and subtracted from front}$$
It seems the front wheels will be enough for traction.
I leave the final steps to you.
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