Understanding the Relationship Between Static and Kinetic Friction on Wet Roads

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Cars are more likely to skid on wet hills than on level roads due to the reduced normal force on an incline, which decreases the static friction available to counteract sliding. On a hill, gravity contributes a component that aids skidding, while on a level road, static friction is stronger and more effective at preventing loss of traction. The normal force is perpendicular to the friction force, and its reduction on a slope leads to a greater likelihood of kinetic friction taking over during a skid. The combination of these factors results in a higher risk of skidding on wet hills compared to flat surfaces. Understanding these dynamics is crucial for safe driving on wet roads.
courtney1121
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Why is your car more likely to skid on a wet hill than on a wet but level road?

Just a somewhat logical guess, but on a level road, the static friction force is much stronger than compared to a wet hill...and you also have accelerationa acting on you opposed to the level road...?
 
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What is the normal force on a hill versus a level road? How is the normal force related to the friction force? Also, on a hill, is there a component of force due to gravity that will assist the skid?
 
well the normal force is going to be less than the gravitational force since acceleration will be inward towards the center of the hill. The normal force will be perpendicular to the friction force. And there should be kinetic friction force that will assit the skid since it's slipping.

Does that make sense?
 
Sounds right to me. Good job.
 
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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