## 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 $$\mu$$s=0.5. Determine the steepest grade (the largest value of the angle of $$\alpha$$) 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.
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$$\alpha$$=0 then NA=30sin$$\alpha$$ (2)0.875*15cos$$\alpha$$+0.475*15sin$$\alpha$$-2.675NA=0 then 13.125cos$$\alpha$$+7.125sin$$\alpha$$-2.675NA=0. Now from (1) and (2) tan$$\alpha$$=7/39 then $$\alpha$$=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$$\alpha$$=72/233 then $$\alpha$$=17.17o

(c)tan$$\alpha$$=0.5 then $$\alpha$$=26.57o
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 Recognitions: Gold Member Homework Help Science Advisor 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.

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 Quote by piacere_space The car weighs 15kN. The coefficient of static friction between the car tires and the road is $$\mu$$s=0.5. Determine the steepest grade (the largest value of the angle of $$\alpha$$) 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$$\alpha$$=0 then NA=30sin$$\alpha$$ (2)0.875*15cos$$\alpha$$+0.475*15sin$$\alpha$$-2.675NA=0 then 13.125cos$$\alpha$$+7.125sin$$\alpha$$-2.675NA=0. Now from (1) and (2) tan$$\alpha$$=7/39 then $$\alpha$$=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$$\alpha$$=72/233 then $$\alpha$$=17.17o (c)tan$$\alpha$$=0.5 then $$\alpha$$=26.57o