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
Messages
15
Reaction score
0
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??
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top