Why Does a Car Behave Erratically When Its Back Brakes Are Locked?

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When a car's back brakes lock, it can cause the rear wheels to skid, leading to a loss of traction and control, resulting in erratic behavior. This is due to the car's weight distribution and dynamics, which are affected more significantly by rear wheel locking than by front wheel locking. In contrast, locked front brakes allow the vehicle to maintain a straighter trajectory because the rear wheels can still steer. Understanding the physics of braking and weight transfer is crucial for explaining this phenomenon. A simpler explanation could focus on how rear wheel skidding disrupts stability and steering control.
shahalam
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I just recently registered with this website and it is incredible! I can't believe how great it is. I stumbled upon here looking for help with a physics question my professor posed to us.

If a car is coating down a hill and locks its front brakes, it will travel in a reasonably straight line. If it locked its back brakes, it would behave erratically. Why?

I really can't think of a reason why this would be. Any help finding the answer would be much appreciated! thanks1 :smile:
 
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Hey man I remember the notes I had on that so they might help a little...

Here is a link:

http://www.mech.uwa.edu.au/DANotes/brakes/vehicles/vehicles.html

Hope that helps a bit
 
thanks a lot big man. is there any other simpler explanation which i could present to my physics professor?
 
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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