Stability of vehicles
When you are travelling around a bend on a motor cycle or in a car the centripetal force is supplied by friction between the wheels and the road surface as you turn into the curve. If you are travelling
too fast for the road conditions, or the condition of your vehicle, one of two things may happen.
1. The friction force between the wheels and the road surface will be insufficient and you will skid. The direction of your skid will be at a tangent to the curve.
2. If the centre of gravity of your vehicle is high it may overturn before starting to skid.
With a motor cycle only the first option is possible. You may of course skid into the kerb and then overturn but it is skidding which will initiate the problem. Cars, buses and trucks are designed so
that even when fully loaded, they should skid before reaching the speed at which overturning would occur. Large heavy items stacked on a roof rack may raise the centre of gravity of a car to
such a height that overturning is a possibility and of course this would also increase the possibility of overturning should the vehicle skid into the kerb.
Although the vehicle is moving, it can be considered to be in a state of dynamic equilibrium. In the vertical direction, the active force is the weight W acting downwards, which is balanced by the upward reactions R1 and R2 of the wheels. The active forces in the horizontal direction are the friction forces F1 and F2 which provide the centripetal force.
The equal and opposite reaction to them is the centrifugal force given by
mv²/r
This method of equating the forces acting on a moving object is known as D’Alembert’s principle
Let the height of the centre of gravity be h.
Let the track width be 2a.
Let the limiting coefficient of friction between the wheels and the road be 𝜇 .
To find the speed at which skidding is likely to occur, equate horizontal forces as the vehicle is about to skid.
centripetal force - centrifugal reaction
F1+F2- mv²/r
In the limit as the vehicle is about to skid
F1 = 𝜇 R1 , F2 = 𝜇 R2
But R1 + R2 = mg, the weight of the vehicle
(1)To find the speed at which overturning is likely to occur, equate moments about the point A in the limit as the vehicle is about to overturn. In this condition, R1 is zero as the nearside wheels are about to lift off the road and all the weight will be carried on the offside wheels.
clockwise overturning moment- anticlockwise righting moment
(mv²/r)h -mg a
(2)Whichever of the equations (1) and (2) gives the lower value of velocity, that will be the limiting value. You should note that both values of limiting velocity are independent of the mass of the
vehicle. Comparison of the two equations shows that
, the vehicle will skid before the overturning speed is reached.
the vehicle will overturn before the skidding speed is reached.
The reaction of the outer wheels will always be greater than that of the inner wheels. To find these reactions at any speed below that at which overturning is likely, take moments about the point A again but this time include the moment of the reaction R1
clockwise overturning moments- anticlockwise righting moment
(mv²/r)h + R1 2a -mg a
Now equate vertical forces to find the reaction of the outer wheels:
R1+R2= mg , R2 = mg –R1
-Properly loaded motor vehicles will skid rather than overturn if their speed around a curve is excessive. Overloading a roof rack can raise the centre of gravity to such a degree that overturning becomes a danger.
-In the condition where there is no tendency for side-slip, the reactions of the inner and outer wheels are equal.
-When a motor vehicle is travelling round a curve, centripetal force is provided by friction between its tyres and the road surface. When skidding occurs, the vehicle continues in a straight line at a tangent to the curve.
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