Centripetal Force + Frictional Force + Braking

[Theorγ]

Homework Statement

Assume that the driver of a car begins to brake his car when the distance from his car to a wall is 107 m. The mass of his car is 1400 kg, and its initial speed prior to braking is 35 m/s. The coefficient of static friction is 0.50, and assume that the car's weight is distributed evenly on the four wheels, even during braking. Based on the listed information, what are the following?:

1. What magnitude of static friction is needed to stop the car just as it reaches the wall?
2. What is the maximum possible static friction f_{s,max}?
3. If the coefficient of kinetic friction is 0.40, at what speed will the driver hit the wall?
4. To avoid the crash, the driver chooses to turn the car so that it just barely misses the wall. What magnitude of frictional force would be required to keep the car in a circular path with the given distance and initial speed?
5. Is the required force less than f_{s,max} so that a circular path is possible?

Homework Equations

a_{c} = v^{2}/r
F_{net} = ma_{c}
f_{s} = m \mu_{s} F_{N}
v^{2} = v^{2}_{0} + 2 a (x - x_{0})

The Attempt at a Solution

I know this is a long problem, but for a majority of the problem, I don't know how to solve. So I'll try to go through each of them as best as I can, and if I get really stuck, I hope you guys can give me hints.

1. For this one, I just used the frictional force equation:
f_{s} = m \mu_{s} F_{N}
f_{s} = 6860 N

2. For this one, I thought that the maximum static friction was the same as above?
f_{s} = m \mu_{s} F_{N}
f_{s} = 6860 N

3. For this one, I used the kinematics equation above in conjunction with the following force equation:

f_{net} = -f_{k}
m a = m \mu_{k} F_{N}
1400 a = 1400*0.40*-9.8
a = -3.92

v^{2} = v^{2}_{0} + 2 a (x - x_{0})
v^{2} = 35^{2} + 2*-3.92(107)
v = 19.6 m/s

4. I have no clue as to what to do for this one, I'll work on this one after I finish the above.

5. I have no clue as to what to do for this one, I'll work on this one after I finish the above.

 

[PhanthomJay]

Homework Statement

Assume that the driver of a car begins to brake his car when the distance from his car to a wall is 107 m. The mass of his car is 1400 kg, and its initial speed prior to braking is 35 m/s. The coefficient of static friction is 0.50, and assume that the car's weight is distributed evenly on the four wheels, even during braking. Based on the listed information, what are the following?:

1. What magnitude of static friction is needed to stop the car just as it reaches the wall?
2. What is the maximum possible static friction f_{s,max}?
3. If the coefficient of kinetic friction is 0.40, at what speed will the driver hit the wall?
4. To avoid the crash, the driver chooses to turn the car so that it just barely misses the wall. What magnitude of frictional force would be required to keep the car in a circular path with the given distance and initial speed?
5. Is the required force less than f_{s,max} so that a circular path is possible?

Homework Equations

a_{c} = v^{2}/r
F_{net} = ma_{c}
f_{s} = m \mu_{s} F_{N}
v^{2} = v^{2}_{0} + 2 a (x - x_{0})

The Attempt at a Solution

I know this is a long problem, but for a majority of the problem, I don't know how to solve. So I'll try to go through each of them as best as I can, and if I get really stuck, I hope you guys can give me hints.

1. For this one, I just used the frictional force equation:
f_{s} = m \mu_{s} F_{N}
f_{s} = 6860 N

No, this part is asking you what force is needed to stop the car, not what is avaiable. Use the kinematic equations and Newton 2.

2. For this one, I thought that the maximum static friction was the same as above?
f_{s} = m \mu_{s} F_{N}
f_{s} = 6860 N

yes, correct, but not the same as above.

3. For this one, I used the kinematics equation above in conjunction with the following force equation:

f_{net} = -f_{k}
m a = m \mu_{k} F_{N}
1400 a = 1400*0.40*-9.8
a = -3.92

v^{2} = v^{2}_{0} + 2 a (x - x_{0})
v^{2} = 35^{2} + 2*-3.92(107)
v = 19.6 m/s

this one is confusing,,,,if the driver locks the brakes immediately and the car skids, then kinetic friction applies...but the driver would be better off applying the brakes without skidding, because of higher friction force available (static), so still , she'll hit the wall, but at less speed than if she locked the brakes.

 

[Theorγ]

1. What magnitude of static friction is needed to stop the car just as it reaches the wall?

v^{2} = v^{2}_{0} + 2 a (x - x_{0})
0^{2} = 35^{2}_{0} + 2 a (107 - 0)
a = -5.72

F_{net} = ma
0 = -F_{brake} - f_{s}
f_{s} = -F_{brake}
f_{s} = -ma
f_{s} = -(1400 * -5.72)
f_{s} = 8008 N

this one is confusing,,,,if the driver locks the brakes immediately and the car skids, then kinetic friction applies...but the driver would be better off applying the brakes without skidding, because of higher friction force available (static), so still , she'll hit the wall, but at less speed than if she locked the brakes.

Does that mean my answer was correct?

 

[PhanthomJay]

Yes, you need 8000 N to stop the car before it hits the wall, but you only have 6860 N available..so the car will hit the wall, unless you swerve...

 

[Theorγ]

okay, then does that mean my answer for #3 is also right?

 

[PhanthomJay]

okay, then does that mean my answer for #3 is also right?

Only if the car locked its brakes and skidded all the way to the wall. The problem is not worded very well...if the driver was able to stop without the tires skidding (ABS Braking system), but just on the verge of skidding, then the speed would be less as she hits the wall. But I guess you are to assume kinetic friction due to the car skidding, but I don't know, but in which case your method is corrrect (I didn't check the math).