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Friction and the difference between rear-wheel, front-wheel and four-wheel

 
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Sep23-09, 02:18 AM   #1
 

Friction and the difference between rear-wheel, front-wheel and four-wheel


The car weighs 15kN. The coefficient of static friction between the car tires and the road is [tex]\mu[/tex]s=0.5. Determine the steepest grade (the largest value of the angle of [tex]\alpha[/tex]) the car can drive up at constant speed if the car has (a) rear-wheel drive, (b) front-wheel drive, and (c) four-wheel drive. Click image for larger version

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My answers are (a)10.18o (b)17.17o (c)26.57o. I just wonder if they are really correct. Thanks!!

Sorry yeah I forgot to post my attempt...

Let the normal forces at the rear wheel and the front wheel be NA and NB respectively. Also let the contact points at the rear and front wheel be A and B respectively.

(a)(1)0.5NA-15sin[tex]\alpha[/tex]=0 then NA=30sin[tex]\alpha[/tex] (2)0.875*15cos[tex]\alpha[/tex]+0.475*15sin[tex]\alpha[/tex]-2.675NA=0 then 13.125cos[tex]\alpha[/tex]+7.125sin[tex]\alpha[/tex]-2.675NA=0. Now from (1) and (2) tan[tex]\alpha[/tex]=7/39 then [tex]\alpha[/tex]=10.18o

Here I assume the rolling friction at the point B is ignorable. (1) is about the equilibrium of all forces in the horizontal direction. (2) is about the equations of moments around B.

Basically I did the same things for (b) and (c).

(b)tan[tex]\alpha[/tex]=72/233 then [tex]\alpha[/tex]=17.17o

(c)tan[tex]\alpha[/tex]=0.5 then [tex]\alpha[/tex]=26.57o
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Sep23-09, 08:51 PM   #2
 
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With a rough calc, they seem to be in the ballpark; if you post your attempt and show how you arrived at these figures, perhaps someone can check the math and method.
Sep25-09, 09:12 AM   #3
 
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Quote by piacere_space View Post
The car weighs 15kN. The coefficient of static friction between the car tires and the road is [tex]\mu[/tex]s=0.5. Determine the steepest grade (the largest value of the angle of [tex]\alpha[/tex]) the car can drive up at constant speed if the car has (a) rear-wheel drive, (b) front-wheel drive, and (c) four-wheel drive. Attachment 20672
My answers are (a)10.18o (b)17.17o (c)26.57o. I just wonder if they are really correct. Thanks!!

Sorry yeah I forgot to post my attempt...

Let the normal forces at the rear wheel and the front wheel be NA and NB respectively. Also let the contact points at the rear and front wheel be A and B respectively.

(a)(1)0.5NA-15sin[tex]\alpha[/tex]=0 then NA=30sin[tex]\alpha[/tex] (2)0.875*15cos[tex]\alpha[/tex]+0.475*15sin[tex]\alpha[/tex]-2.675NA=0 then 13.125cos[tex]\alpha[/tex]+7.125sin[tex]\alpha[/tex]-2.675NA=0. Now from (1) and (2) tan[tex]\alpha[/tex]=7/39 then [tex]\alpha[/tex]=10.18o

Here I assume the rolling friction at the point B is ignorable. (1) is about the equilibrium of all forces in the horizontal direction. (2) is about the equations of moments around B.

Basically I did the same things for (b) and (c).

(b)tan[tex]\alpha[/tex]=72/233 then [tex]\alpha[/tex]=17.17o

(c)tan[tex]\alpha[/tex]=0.5 then [tex]\alpha[/tex]=26.57o
Sorry for not responding sooner.
I agree exactly with your method and answers, nice work! Intuitively, one would expect an all wheel drive vehicle to handle a grade steeper than front wheel or rear wheel drive only. But between front and rear wheel drive, you've got to crunch out the numbers to see which amongst the two can handle the steeper grade, because it depends on the location of the car's center of gravity. You've shown that quite nicely.
Jul10-11, 08:53 PM   #4
 

Friction and the difference between rear-wheel, front-wheel and four-wheel


hello. i have a question. How do you do to find these numbers i dont understand so much.. 72/233.... its the answer of question B.. thank you so much.
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