The picture above shows a skier making a right turn by leaning to the right and transferring his/her weight on to the right-hand ski.
The pictures above show a motorcyclist turning right by leaning right and a plane turning left by leaning left. How can a plane lean over? This is achieved by flaps near the ends of the wings called ailerons.
The first picture (immediately above) shows an aileron in its neutral position; the middle picture shows it raised and the third picture shows it lowered. When the aileron is raised, the drag over the upper wing surface increases because M, defined in post 17.17 is increased (the upper surface is less streamlined – see post 17.18). So, relative to the air, the wing moves more slowly (see post 17.17). Relative to the wing, the speed of the air moves more slowly (post 16.4) over its upper surface, so the lift is reduced (post 17.16). When the aileron is lowered, the drag over the lower wing surface is increased, so the air moves more slowly over the lower wing surface – increasing the lift (post 17.16).
A plane can be made to lean to the left by raising its right wing (lowering its right-hand aileron) while simultaneously lowering its left wing (raising its left-hand aileron).
So why does leaning to the left make a plane turn left? The picture above shows the lift force, F, acting upwards perpendicular to the wings. If the wings are tilted by an angle θ to the horizontal, this force has a component (post 16.50) Fsinθ in the horizontal direction, as shown. This horizontal component of the lift acts as the centripetal force (post 17.13) to turn the plane.
The magnitude of the centrifugal force is given by mω2r where m is the mass of the object (in this case the plane), ω is the angular speed and r is the radius of the turning circle (post 17.13). Since ω is related to linear speed by v = ωr (post 17.12), mω2r = mv2/r. If it is the horizontal component of the lift that provides this centripetal force, we can write that
Fsinθ = mv2/r.
Multiplying both sides of this equation by r and dividing mv2 shows that the radius of the turning circle provided by tilting the plane, through an angle θ, is
r = mv2/Fsinθ.
The same ideas apply to an object (like the skier or the motor-cyclist) that is moving along the ground. Gravity exerts a force of magnitude mg, where m is the mass of the object and g is the magnitude of the earth’s gravitational field, that tends to pull the object into the ground (post 16.16). According to Newton’s third law of motion (post 17.26), the ground then exerts an equal and opposite force (sometimes called the ground reaction force). When the object leans, the horizontal component provides the centripetal force to turn it.
However, if our object is a road vehicle, a well-designed road is not flat when it turns a corner. In the picture above, A is higher than B to assist a cornering vehicle. This leads to a tilt of the road surface called camber.
In the picture above, the vehicle rests on a slop with a tilt angle of θ. Gravity exerts a downward vertical force of mg on the vehicle (post 16.16). Since the road surface is tilted by an angle θ with respect to the horizontal, a line perpendicular to this surface is tiled by θ with respect to the vertical, as shown in the picture. So, the component of the gravitational force acting perpendicular to the road surface is mgcosθ (post 16.50). The picture shows the ground pushing up on the vehicle with an equal but opposite force. This force has the same effect as F in the picture of the forces acting on a plane that is turning.
There are two potential problems for a vehicle on a camber. One: if θ is too big, the vehicle will fall over (post 17.22). Fortunately, the road can be designed (using the equation given above) so that this is unlikely to happen for well-designed vehicles. Two: what stops the vehicle from sliding down the slope? The force that opposes this motion is friction (post 16.19). In the next post, we will see how to ensure that sliding doesn’t happen.