Car traveling on horizontal road with force of static friction

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SUMMARY

The discussion focuses on calculating the stopping distance of a car traveling at 50 mi/hr under two conditions: wet roads with a coefficient of static friction (μs) of 0.100 and dry roads with μs of 0.600. The equations of motion and Newton's second law are utilized to derive the stopping distances. The net force acting on the car is equal to the force of static friction, which is critical for determining the deceleration and, subsequently, the stopping distance.

PREREQUISITES
  • Understanding of Newton's second law (F_net = ma)
  • Knowledge of static friction and its coefficient (μs)
  • Familiarity with kinematic equations of motion
  • Basic concepts of forces acting on a vehicle in motion
NEXT STEPS
  • Calculate stopping distances using different coefficients of static friction
  • Explore the impact of initial velocity on stopping distance
  • Study the effects of road conditions on vehicle dynamics
  • Learn advanced kinematic equations for motion analysis
USEFUL FOR

Students studying physics, automotive engineers, and anyone interested in vehicle dynamics and safety calculations.

Color_of_Cyan
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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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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.
 
That helped a lot, thanks!

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

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