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

[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.

My answers are (a)10.18^o (b)17.17^o (c)26.57^o. 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 N_A and N_B respectively. Also let the contact points at the rear and front wheel be A and B respectively.

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

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 $$\alpha$$=17.17^o

(c)tan$$\alpha$$=0.5 then $$\alpha$$=26.57^o

 

[PhanthomJay]

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.

 

[PhanthomJay]

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. View attachment 20672
My answers are (a)10.18^o (b)17.17^o (c)26.57^o. 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 N_A and N_B respectively. Also let the contact points at the rear and front wheel be A and B respectively.

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

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 $$\alpha$$=17.17^o

(c)tan$$\alpha$$=0.5 then $$\alpha$$=26.57^o

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.

 

[raquelmachin6]

hello. i have a question. How do you do to find these numbers i don't understand so much.. 72/233... its the answer of question B.. thank you so much.

 

[DeeNos]

Overall, your calculations and answers seem to be correct. You have correctly used the equations for equilibrium and moments to find the maximum angle at which the car can drive up a slope at constant speed for each type of drive. Good job!