Solving Friction Problems: Why Only Static Friction Applies?

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The discussion centers on understanding why only static friction applies when calculating the stopping distance of a car traveling at 50 mi/h on a rainy highway with a static friction coefficient of 0.10. Static friction prevents relative motion between the tires and the road until a threshold is reached, which is crucial for maintaining control while braking. The conversation clarifies that as long as the tires do not skid, static friction is the force at play, particularly in scenarios involving anti-lock braking systems (ABS). The distinction between skidding and rolling is emphasized, reinforcing the importance of static friction in stopping distances. Overall, the key takeaway is that static friction is essential for effective braking without losing tire grip.
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The question is:

A car is traveling at 50 mi/h on a horizontal highway. If the coefficient of static friction between road and tires on a rainy day is 0.10, what is the minimum distance in which the car will stop?

I know how to solve it using static friction, but I don't understand why only static friction is needed in the problem, and kinetic friction does not apply.
 
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What is static friction? A force which prevents any relative motion until its threshold is reached. So in order to keep the friction between the road and your tires, you must maintain a minimum acceleration. And once you find this acceleration, you can find the minimum distance it will take to stop.
 
so does that mean, when I press the breaks, there is only static friction between the tires and the road?
 
With (an ideal) ABS, I suppose, there's only static friction(?)
 
That is correct, as long as your wheels are not skidding they maintain static friction with the road. Consider the difference between skidding and rolling.
 
ohh, ok, I get it now. Thank you for the help
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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