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Bike/motorcycle leaning in a curve. What happens to the normal force N? 
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#1
Nov312, 06:35 PM

P: 374

Hello Forum,
If a bike is negotiation a curve, the static friction force will supply the centripetal force F_c. If we go too fast and the friction is not enough we skid along the tangent. There are two forces: the weight W=mg pointing down, the normal (contact) force N pointing up and equal to the weight W, and the static friction 0<=f_s<= mu_s*(N)= m_s*(mg). If we go fast, we automatically lean into the curve. Why? Some free body diagrams show the normal force being inclined w.r.t the pavement and pointing along the rider axis. How can that be? The normal force is called normal because it is perpendicular to the floor. Does the leaning increase the normal force somehow? If we increase N we automatically increase f_s and supply a larger centripetal force. Maybe there is "camber thrust": the normal force gets inclined because there is a deformation of the tires..... Or does the leaning simply help creating a torque about point of contact due to the weight W that counterbalance the "centrifugal" force torque (centrifugal force does not exist really, only in noninertial frames).... Otherwise, how come the bike does not fall if the CM is outside the base of support (wheels)? Thanks, fisico30 


#2
Nov312, 07:48 PM

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P: 11,747

Let's analyze the situation in the inertial reference frame of the ground. There are two forces that matter, both acting on the bottom of the bike tire:
1. A sideways frictional force points towards the center of the circular path (centripetal force). Let's say the center of the circle is to your left, then this force is to your left. 2. The upward normal force is equal in magnitude to the weight (gravitational force) of you and the bike. Suppose you try to stay upright, without leaning. Let's look at the torques produced by both forces, using the center of mass of you and the bike as the center of rotation. 1. The frictional force produces a clockwise torque, as seen from the rear of the bike. 2. The normal force produces zero torque, because it acts along a line that passes through the center of mass. Net result: a clockwise torque which makes you fall over to your right. Now suppose you lean to your left. 1. The frictional force still produces a clockwise torque. 2. The normal force no longer acts along a line through the center of mass. It produces a counterclockwise torque. When you lean at a certain angle, the two torques are equal in magnitude and opposite in direction, giving zero net torque. This angle depends on the magnitude of the frictional force, which in turn depends on your speed going around the curve. 


#3
Nov312, 08:04 PM

P: 7

Can the normal force during the turn, be figured as ((W)(sine(theta))



#4
Nov312, 08:29 PM

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P: 11,747

Bike/motorcycle leaning in a curve. What happens to the normal force N?
The upward normal force always equals W (=mg). It has to, otherwise you'd sink straight down into the road or levitate upward off the road. (Unless there's another vertical force acting on the bike besides the normal force and gravity.)
If θ is the angle between the bike and the vertical direction, then W sin θ is the component of N perpendicular to the bike, that is, the component that produces a counterclockwise torque in the setup described above. 


#5
Nov312, 11:49 PM

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#6
Nov412, 09:05 AM

P: 374

Hello Jtbell,
thanks for your inputs. I agree with you but I have been corrected on the following points: So, from the ground (inertial frame) there are only two forces: gravity W (pointing down), the normal force N (pointing up and perpendicular to the floor) and the static friction force f_s pointing radially inward. I thought, like you, that the leaning of the bike or motorcycle has the only purpose to counteract the clockwise torque produced by the static friction about the point of contact. It is all about stability. The leaning does not increase the centripeta force whose max value is mu_s (N). But some told me about the camber thrust: the tire gets deformed and provides some extra radially inward force to add to the friction force. Also, some draw the normal force N at an angle with respect to the vertical....Is that incorrect? Normal means perpendicular. But does the normal force became oblique due to camber thrust which increases the normal force value, hence the static friction, hence the centripetal force? thanks fisico30 


#7
Nov412, 01:30 PM

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P: 7,109

I'm assuming a flat (nonbanked) road in my post.
The usual usage of the term normal force refers to a vertical force related to gravity. During transitions in lean angle where the center of mass of the motorcycle and rider is accelerating, the normal force can change temporarily, lesser if the center of mass is accelerating downwards, greater if the center of mass is accelerating upwards. Camber thrust is one of the components of the centripetal force that causes the motorcycle to follow a curved path. The predicted radius based on camber thrust is much smaller than the actual radius of a cornering motorcycle, so the effect is not large, at least not for a motorcyle. If you create a rig with two cone shaped "tires", one in front of the other (like a motorcycle), and if the axis of the two "cones" is parallel, the rig will tend to move in a nearly straight line, with a lot of slippage across the contact patch of the cones. The more common explanation is that the front tire is turned "inwards" of the rear tire (once a stable lean is established). If you draw imaginary line extensions from the axis of the front and rear tire, they will cross at some point below the surface of a road. If you then have a vertical line from where those axis lines cross back to the surface of a road, then where that vertical line meets the surface of the road will be the approximate center of the circular path that the motorcycle follows. Due to deformation at the contact patch of the tires, the actual radius will be slightly larger. 


#8
Nov512, 01:37 PM

P: 374

Thanks rcgldr!
In an old post, you said: "With a car of the same weight, the larger the contact patch area, the lower the load per unit area of the contact patch, which translates into somewhat higher grip, up to a point, due to something called tire load sensitivity, where the coefficient of friction decreases the with amount of load. There's a point where increasing contact patch area beyond some near optimal size has little additional effect." 1) So a large contact patch result in a lower pressure (load per unit area). Why does that result in a higher grip? Are you saying that after a certain contact patch area we start losing friction (traction)? Is it because the pressure becomes too low? 2) in regards to a bike/bicycle leaning into a curve, just to keep things straight: From the net: Cornering Force Opposing lateral force is cornering force that the tyres create when you turn the wheel into a corner. By completing the turn, cornering force has overcome lateral force. The important thing here is that tyres, and only tyres, can generate cornering force, suspension systems can only affect how tyres generate and share that cornering force. Camber Thrust The second way the tyres generate cornering force is by camber thrust. Positive camber is when the top of the wheel leans outwards; negative camber is when the top of the wheel leans inwards. Camber thrust is the force that moves the tyre in the direction it is leaning. Like slip angle, the greater the camber angle, the greater the camber thrust (cornering force) generated in the direction the tyre is leaning. So, does the normal force go from being vertical to being at an angle? If so, what is the normal force perpendicular to in that case? Does the action of leaning the bike increase the necessary centripetal force? How? How many forces do we have? Weight W, normal force N, static friction f_s and camber thrust? thanks fisico30 


#9
Nov512, 05:08 PM

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wiki_tire_load_sensitivity.htm The wiki article mentions that maximum horizontal force is proportional to the normal force raised to somwhere from .7 to .9 power. If there was no load sensitivity, then it would be just the normal force (or normal force raised to 1.0 power). There's also the issue of heat dissipation, and having a larger tire (taller or wider) provides more surface area for heat dissipation. For a high downforce race car, such as Formula 1, the camber is set a bit more negative, so that the tire's inside temperatures are a bit higher than the outside temperatures. I'm not sure if the reason for this is to produce camber thrust as opposed to other reasons for the extra bit of negative camber setting. 


#10
Nov612, 07:44 PM

P: 374

Thanks rcgldr,
"The load sensitivity of most real tires in their typical operating range is such that the coefficient of friction decreases as the vertical load, Fz, increases. The maximum lateral force that can be developed does increase as the vertical load increases, but at a diminishing rate." So we have two types of grip: grip in the linear, forward direction (good for acceleration and breaking) and lateral grip (good for stability in curves). Why do you think the the forward grip decreases with increase in the load? Conceptually, I would think it would increase since more contact force is developed..... fisico30 


#11
Nov612, 08:32 PM

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#12
Nov712, 10:26 AM

P: 374

Thanks again rcgldr.
I guess I was thinking that the coefficient of friction must always increase as more weight (more load) is applied, since the stickiness increases.... One last comment about camber thrust: as the wheel leans (negative camber), the tire deforms in such a way that it is AS IF the rim was on top of a mini banked curve provided by the deformed tire. That way, the normal force is now oblique to the floor, the same way it is when a car is on a banked curve. Sure, the normal force between the tire and the floor is still vertical..... The increase in grip due to camber thrust derives from an increase in normal force due to the tire deformation upon leaning.....That increase in normal force leads to an increase in static friction force, which provides a larger centripetal force to keep the bike from sliding out of the curve... thanks fisico30 


#13
Nov812, 04:14 PM

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P: 7,109

Archived article (so the images are gone), disputing camber thrust as a major contributor to conering on a bike:
bike1.htm From another article: When cornering at an angle of 45° and 70mph. the turn radius will be 327 ft. and at half that speed, 35mph., the radius will be a quarter of that or 82 ft.  but as shown earlier the cone radius will be only 1.5 ft. tyres.htm Camber thrust effect should slightly reduce the amount of steering input required to make a turn, while deformation (twisting) at the contact patch results in a slip angle that slightly increases the radius of a turn for a given steering input. The slip angle at the front and rear tire may be different, and I'm not sure of the combined effects versus steering input required to hold a turn. 


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