Rear Wheel Driven Car Rolling Friction

AI Thread Summary
In a rear-wheel-driven car, the friction on the rear wheels acts forward to accelerate the vehicle, while the friction on the front wheels acts backward due to their role in maintaining stability. The direction of rolling friction for the car as a whole is forward, opposing any slipping between the tires and the road. The rear wheels typically experience a greater magnitude of friction because they are responsible for propulsion. Friction always opposes slipping, which means it acts in the direction of the car's motion during acceleration. Understanding these dynamics is crucial for analyzing vehicle performance during acceleration.
pcdagr8
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if a rear wheel driven car is accelerated
i think...
1. on rear wheels friction is forward
2 on front wheels it is backward


my questions-
1.wats d direction of rolling friction on the car as a whole?(is dis question correct?)

2.which wheel will have greater magnitude of friction??
 
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friction is always in direction opposite that of motion.
 
denverdoc said:
friction is always in direction opposite that of motion.

going by your logic
You mean in pure rolling motion the friction should act backward??
 
friction opposes slipping between surfaces

A better way to describe friction is to say that it always opposes slipping between surfaces. Consider the rear tires. Without friction, they would just spin, slipping to the rear with respect to the road. Friction with the road opposes that slipping by acting towards the front. The friction accelerates the car forward, so friction and the car's motion are in the same direction.
 
Thanks Al, a nice explanation. I was only considering situation where there was already angular velocity and see now problem asks during acceleration. My bad.
 
Thanks Al !:smile:
but u haven't answered my second question...
please see if u can tell me something about it.
 
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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