# Nature of frictional force

I was thinking about a situation where a person in standing on the ground with some friction. The frictional force is directly proportional to the normal reaction acting on him by the ground. Assume that he leaned forward i.e his center of mass is not vertically up the point of contact of him with the ground.Then, does the frictional force change?

My thoughts:Although the person has leaned forward the force $mg$ acting on him will be vertically downward. Since normal reaction is always normal to the surface of contact its magnitude will remain same and thus frictional force will remain same.But, I have seen athletes starting the running race with a leaning forward position which would be mostly for increasing the friction between their shoes and the track. So, I'm in a dilemma. Please help.

It's not correct that runners lean forward to begin a race in order to increase friction. They lean forward because otherwise, they would experience no propulsion whatsoever because static friction is zero when the runner is completely upright.

When the runner leans forward and flexes his leg muscles, he exerts a horizontal force on the track in the backward direction. The track responds by exerting an equal and opposite frictional force (unless there is slipping) on the runner in the forward direction that propels him forward.

Generally speaking, the more a runner leans forward at the start, the larger the horizontal component of the force exerted by his legs against the ground, and the larger the frictional force he will experience. As a result, his initial acceleration will be greater.

When the person is standing perfectly upright, the frictional force on him is actually zero and the normal reaction is $mg$. The frictional force is the horizontal component of the contact force and normal reaction is the vertical component. Therefore, unless there is a horizontal force, there will not be any friction whatever may be your composure.
In the second case when your COM moves forward, the normal reaction now tries to prevent toppling over by shifting its position and the line of action so that it can balance torques about the centre of mass. This is possible only in certain positions. Try to see the dynamics of toppling over to understand how normal reaction tries to balance the torque of weight acting at centre of mass, and under what conditions can it be balanced.
In case of athletes, balancing and optimizing their composure for best configuration of COM is very important, though I don't think that bending forward has anything to do with reduction or increasing the friction.(but rather to optimize Aerodynamic and classical forces)

• Then, does leaning forward affect the acceleration of a person in any way? – Rajath Krishna R Oct 6 '13 at 4:41
• It does but only by secondarily changing the position of normal reaction and by shifting the centre of mass and thereby changing the dynamics of it all. Moreover, the resistive air drag is also modified. To understand in detail, you will need to look at a detailed account of bio-mechanics and dynamics of human body, may be in a good book on sports physiology. – Satwik Pasani Oct 6 '13 at 5:00

If the center of mass and the point of contact are not vertically aligned, the person will tend to rotate; to see this, take the torque about the contact point. Hence your model is incomplete.

The athlete can only be in static equilibrium (before the start of the race) if he contacts the ground at two points, namely feet and hands. Try working out whether the friction is greater in such a model.