Car traveling on horizontal road with force of static friction

In summary: I thought it was supposed to be the other way around.You are correct. The equation should be F_net = ma - fs, where F_net is the net force in the direction of the acceleration.
  • #1
Color_of_Cyan
386
0

Homework Statement


A car is traveling at 50 mi / hr on a horizontal highway

a. If the coefficient of static friction between the road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop?

b. What is the stopping distance when the surface is dry and μs = 0.600?

Homework Equations


∑ f = ma

fs = (μs)(normal force)

Xf = Xi + (Vi)(t) + (1/2)at2


The Attempt at a Solution


Do not know where to start, please help.

Part a:

∑ Fx = ma - fs = 0

fs = (μs)(normal force)

∑ Fy = (normal force) = mg

ma = fs therefore ma = (μs)(normal force)

a = (μs)(normal force) / mass

Now: vf = 0

vi = 73.3 ft / second

Xf = Xi + Vi t + (1/2)at2

0 = 50 mi / hr + [ (0.100)(normal force) / mass ][t2]
 
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  • #2
Color_of_Cyan said:

Homework Statement


A car is traveling at 50 mi / hr on a horizontal highway

a. If the coefficient of static friction between the road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop?

b. What is the stopping distance when the surface is dry and μs = 0.600?

Homework Equations


∑ f = ma

fs = (μs)(normal force)

Xf = Xi + (Vi)(t) + (1/2)at2


The Attempt at a Solution


Do not know where to start, please help.
Actually, you have started off pretty well.
Part a:

∑ Fx = ma - fs = 0
Yes, but try to avoid the use of pseudo inertial forces, where practical, when using Newton's 2nd law. Rather use F_net = ma, where F_net in this case is fs, in the direction of the acceleration. It's the same result as F_net = ma - fs = 0, but it avoids confusion down the road.
fs = (μs)(normal force)

∑ Fy = (normal force) = mg

ma = fs therefore ma = (μs)(normal force)

a = (μs)(normal force) / mass

Now: vf = 0

vi = 73.3 ft / second

Xf = Xi + Vi t + (1/2)at2

0 = 50 mi / hr + [ (0.100)(normal force) / mass ][t2]
Rather than use this kinematic equation that relates distance, acceleration, initial velocity, and time, use the kinematic equation of motion that relates distance, initial and final velocity, and acceleration. That avoids solving for the time.
 
  • #3
That helped a lot, thanks!

I didn't know you were supposed to set netforce equal to force of static friction.
 

1. What is the force of static friction on a car traveling on a horizontal road?

The force of static friction on a car traveling on a horizontal road will vary depending on the weight of the car, the surface of the road, and the speed of the car. However, the maximum force of static friction that can be exerted on the car will be equal to the coefficient of static friction multiplied by the weight of the car.

2. How does the force of static friction affect the motion of the car?

The force of static friction acts in the opposite direction of the car's motion. This means that it helps to keep the car from slipping or sliding on the road. It also helps to maintain the car's speed and direction while it is in motion.

3. What factors can affect the force of static friction on a car?

The factors that can affect the force of static friction on a car include the weight of the car, the surface of the road, the speed of the car, and the coefficient of static friction between the car's tires and the road surface. Other factors such as the weather conditions and the condition of the tires can also play a role.

4. How does the coefficient of static friction impact the force of static friction on a car?

The coefficient of static friction is a measure of the amount of friction between two surfaces in contact. The higher the coefficient of static friction, the greater the force of static friction that can be exerted on the car. In other words, a higher coefficient of static friction means that the car will have a better grip on the road and will be less likely to slip or slide.

5. Can the force of static friction ever be greater than the weight of the car?

No, the force of static friction can never be greater than the weight of the car. This is because the maximum force of static friction that can be exerted on the car is equal to the coefficient of static friction multiplied by the weight of the car. If the force of static friction were to exceed the weight of the car, the car would start to accelerate in the opposite direction of its motion, causing it to slip or slide on the road.

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