Further Applications of Newton's Law : Friction, Drag, and Elasticity

[EnricoHendro]

Homework Statement

Calculate the maximum acceleration of a car that is heading up a 4o slope (one that makes an angle of 4o with the horizontal) under the following road conditions. Assume that only half the weight of the car is supported by the two drive wheels and that the coefficient of static friction is involved—that is, the tires are not allowed to slip during the acceleration. (Ignore rolling.) (a) On dry concrete. (b) On wet concrete. (c) On ice, assuming that μ s = 0.100 , the same
as for shoes on ice

Relevant Equations

a=(g.sinθ+μs.g.cosθ)/2

θ = 4
μs = 1
Fnet = Wpararell + fs
m.a = 1/2.m.g.sinθ + μs.1/2.m.g.cosθ (divide by m)
a = (g.sinθ+μs.g.cosθ)/2
a = ((9.8)sin4 + (1)(9.8)cos4)/2
a = 5.23 m/s^2

hello guys, I'm having trouble with this problem. Can anybody help me correct my attempt and explain it to me?? thanks

 

[Abhishek11235]

Please elaborate your question and use Latex. Apart from the fact that you have taken friction due to 1 tire,I don't see any other error

 

[EnricoHendro]

Hello there,
I found this solution from Database-Physics-Solutions. In this solution, it uses F=f-Wx, instead of F=f+Wx.
now my question is why do they use F=f-Wx instead of F=f+Wx?? I mean, the friction should act against the movement of the car right?? (the car is going upward and the Wx is in same direction as friction. Shouldn't it be f+Wx instead??) My second question is why is the f = 1/2 μs.mg Cosθ while the Wx = mg Sin θ ?? I mean since only half of the weight of the car is being supported by the 2 drive wheels??
Sorry for my bad english.
And by the way, what is Latex?? I don't know what latex is, sorry :)

 

[haruspex]

the friction should act against the movement of the car right?

No. see section 2 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/

why is the f = 1/2 μs.mg Cosθ while the Wx = mg Sin θ ?? I mean since only half of the weight of the car is being supported by the 2 drive wheels?

Only half the weight is on the drive wheels, so the combined normal force on those is 1/2 mg Cosθ. Friction stops those spinning and leads to the frictional force up the slope. This has a max of 1/2 μs.mg Cosθ (it could be less in general). There is no frictional force on the other wheels, so the total frictional force acting on the car up the slope is 1/2 μs.mg Cosθ.
The mg Sin θ acts down the slope on the car as a whole.