Friction Forces on 2WD vs 4WD Vehicles

[danielsmith123123]

Homework Statement

What is the maximum acceleration a car can undergo if
the coefficient of static friction between the tires and the
ground is 0.8? Assume (i) 2-wheel drive, (ii) 4-wheel drive.
[Hint: Consider the normal force on each tire, assumed
equal.]
(a) 0.4 g, 0.4 g.
(b) 0.4 g, 0.8 g.
(c) 0.8 g, 0.4 g.
(d) 0.8 g, 0.8 g.
(e) Cannot tell without more information.

Relevant Equations

N mu = F friction

My answer for this question is d as every car has the same result for the force of friction since the normal and coefficient of static friction is the same. I cannot find an answer online so can anyone help verify this? Thank you.

 

[PeroK]

Homework Statement: What is the maximum acceleration a car can undergo if
the coefficient of static friction between the tires and the
ground is 0.8? Assume (i) 2-wheel drive, (ii) 4-wheel drive.
[Hint: Consider the normal force on each tire, assumed
equal.]
(a) 0.4 g, 0.4 g.
(b) 0.4 g, 0.8 g.
(c) 0.8 g, 0.4 g.
(d) 0.8 g, 0.8 g.
(e) Cannot tell without more information.
Relevant Equations: N mu = F friction

My answer for this question is d as every car has the same result for the force of friction since the normal and coefficient of static friction is the same. I cannot find an answer online so can anyone help verify this? Thank you.

Why (d)? What about a 0-wheel drive?

 

[kuruman]

My answer for this question is d as every car has the same result for the force of friction $\dots$

Exactly what result is this that is the same? Consider the simpler case of a unicycle and a bicycle. Both are 1-wheel drives. As far as calculating the maximum acceleration in each case, what parameters are the same and what are different? Assume that the same person rides either one and that the vehicles have the same mass.

 

[jack action]

My answer for this question is d as every car has the same result for the force of friction since the normal and coefficient of static friction is the same.

With that only statement, how do you know the answer is not a), where the answer is also the same acceleration for both configurations?

You are missing one relevant equation, one that should link the acceleration to the friction forces involved.

Just so we are clear on definitions:

For four-wheeled vehicles [...], this term [two-wheel drive (2WD)] is used to describe vehicles that are able to power at most two wheels, [...]

4WD
Four-wheel drive (4WD) refers to vehicles with two axles providing torque to four axle ends.

 

[berkeman]

Homework Statement: What is the maximum acceleration a car can undergo if
the coefficient of static friction between the tires and the
ground is 0.8? Assume (i) 2-wheel drive, (ii) 4-wheel drive.
[Hint: Consider the normal force on each tire, assumed
equal.]
(a) 0.4 g, 0.4 g.
(b) 0.4 g, 0.8 g.
(c) 0.8 g, 0.4 g.
(d) 0.8 g, 0.8 g.
(e) Cannot tell without more information.
Relevant Equations: N mu = F friction

My answer for this question is d as every car has the same result for the force of friction since the normal and coefficient of static friction is the same. I cannot find an answer online so can anyone help verify this? Thank you.

Have you ever been to a drag race? Why don't pavement dragsters use 4WD?

And that "hint" is silly...

[Hint: Consider the normal force on each tire, assumed equal.]

 

[WWGD]

Have you ever been to a drag race? Why don't pavement dragsters use 4WD?

And that "hint" is silly...

[Hint: Consider the normal force on each tire, assumed equal.]

View attachment 335873

Drag race? Can they run wearing all those high heels? ;).

 

[Lnewqban]

[Hint: Consider the normal force on each tire, assumed
equal.]

N mu = F friction

... every car has the same result for the force of friction since the normal and coefficient of static friction is the same.

Each tire is supporting $$ \frac {m_{car}g} {4}$$ Therefore, that should be the magnitude of the normal force at each contact patch.
Assuming infinite power delivered to each wheel, four versus two contact patches would be able to generate a total pushing force of ...