1. Moments and Stability
Elliott

2. Stability Issues Going Around Corners

3. This bus has a low centre of mass and a wide track (distance between the wheels)
You can see that there is a line of action of the weight that acts vertically downwards from the centre of mass.

4. Which Direction is The Moment?

5. The bus has its centre of mass half-way between the wheels
The bus has its centre of mass half-way between the wheels.  The distance from the centre of mass to the tyre is d metres, and its height above the road is h metres.  If the mass of the bus is m kilograms, its weight is mg newtons.

6. Suppose we tip the bus over by an angle of θ to the road:
Moment (anticlockwise) = mg × d sin θ

7. Critical Point
There is a critical point at which the bus might fall back or tip over.  This is where the centre of mass is directly above the point of contact of the tyre with the road. Point P is vertically  above point Q.

8. Calculating the Critical Point
If we know what distance d is and the height h are, we can easily work out the angle at which the bus will tip over.  The distances are shown in the diagram.

9. Check Point
A lorry has a mass of kg.  Its track (width between the wheels) is 2.0 m, and its centre of mass is 0.75 m above the road surface.
It is travelling due North.  A cross-wind from due West is acting on the side of the lorry.  This is acting on the lorry with a force of 5000 N, and the line of action of the force is 3.0 m above the road surface.
 (a) Show that the maximum angle to the vertical that the lorry could tilt before it tips over is about 37o.
 (b) Calculate the moment made by the wind, hence the angle to which the lorry would tilt.
 (c) Will the lorry tip over?

10. Answer (a) h = 0.75 m and d = 1.0 m (centre of mass is in the middle)
tan  θ  = h/d = 0.75/1.0 = 0.75
θ = tan-1 (0.75) = 36.9 o
But this is from the horizontal.
Angle from the vertical = = 53.1o

11. (b) Moment = mg × d sin θ = 10000 × 9. 8 × 0. 75 sin θ = 98000 N × 0
(b) Moment = mg × d sin  θ  = 10000 × 9.8 × 0.75 sin  θ  = N × 0.75 sin  θ
Moment from the wind = 5000 N × 3.0 m
= N m
15000 = 98000 N × 0.75 sin  θ
 sin  θ  = ÷ (98000 × 0.75) = 0.204
θ = sin-1 (0.204) = 11.5o (to the horizontal)

12. Tip The Bus Further…
The line of action of the weight is to the outside of the tyre, so the turning moment is clockwise.  The bus tips over on its side.

13. Moment = mg × d sinθ

14. Stability in Aeroplanes
When an aeroplane flies, there is a centre of lift, which is an imaginary point through which all the lift from the wings appears to act. Ideally the centre of mass of the aeroplane should be directly underneath this. Since the line of actions of both forces coincide, there is no turning moment, and the aeroplane is stable.

15. Adjusting Lift
When the pilot uses the elevators (flaps at the tail to make the aeroplane move up and down), a force is applied to the tail, causing a turning moment to act.

16. Turning
When the pilot wants to turn, he uses the ailerons (flaps at the end of each wing).  One aileron is raised, and the other is lowered, to make a couple.  The aeroplane rolls.

17. Check Your Progress
The pilot can take petrol from both tanks, and he selects which tank to use. He needs to keep both tanks balanced, by using 10 litres from one tank, then 10 litres from the other tank. Every few minutes he needs to change over the tanks. Suppose he forgets to change over, and takes almost all the fuel from the right tank. How do you think this will affect the handling of the plane? Explain your answer.

18. There will be a turning moment to the left as the right wing is much lighter than the left.  The pilot would have to steer to the right to raise the left wing.
At best this will make the aeroplane much more difficult to handle.  As the pilot slowed down to land, there is a strong chance that the left aileron (a flap which raises the left wing) would become less effective, so the wing dropped right down, causing an uncontrollable spin.