# Confused on Static Friction - Walking and Driving

indoguy427
Hey Guys,

This is my first time posting ... sooo here it goes!

I have spent the past hour trying to figure out how we walk (in a physics sense) - I can't seem to find anything good on google that really answers my question.

I have a general idea that when i put my foot down, I push backwards against the floor and that causes the floor to push me forward. This is a very generic description though and I am trying to decompose the matter into FBD's.

So this is my understanding thus far. When I wish to walk forward I plant my foot on the floor. Then I use my muscles to apply a force to my foot backwards. If there was no on the surface, the force applied to my foot would cause it to accelerate in a backwards direction and thus cause me to slip. Because there is friction, the force of static friction acts on my foot in the forward direction, balancing the force I applied. This causes my foot to be stationary.

So here's my point of confusion...where does the force from the ground pushing back on me come from? All I can see here are two forces acting on my foot: 1.) my leg pushing my foot backwards 2.) static friction pushing my foot forwards, resulting in a 0 net force on the foot.

If someone out there is listening, can you please explain how we are able to walk in terms of individual forces, so I can understand the picture completely?

Also, I have one more question and that relates to turning a car. I was looking at an example in my physics book and it was describing a car going around a turn (uniform circular motion) on a road with certain friction coefficient. In the FBD, they only drew 3 forces: gravity, normal, and static friction force pointing into the turn (this was the centripetal force). I was confused as to how static friction came into play at all in this. Also, I don't understand how static friction acts alone in the horizontal direction. My understanding of static friction was that, when you apply a force to an object, static friction force appears and resists motion, upto the slip limit. For example, if a block is sitting on the floor, there is no static friction force. If I apply a force to the box, then there is a force of static friction to oppose my push. However, in this FBD in my book, there was just one horizontal force and that was the static friction force. I don't see how it would arise unless I was pushing the car laterally.

Can someone please help me understand how the car is able to turn with a physically rigorous description (including forces exerted by the tire, etc.) because I am having the woorrst time trying to understand it!

Thanks a lot for your help guys... really appreciate it!

Homework Helper
For the walking case, you apply a backwards force on the earth, which can only happen if the surface of the Earth is applying an equal and opposing forward force. (You can't push something unless it pushes back). Since neither you or the Earth are constrained, both you and the Earth are accelerated by this force, but since the Earth is so massive, you experience almost all of the acceleration, in a forwards direction, and the Earth is translated and rotated a very tiny amount in the backwards direction. If you use the Earth as a frame of reference, then you do all of the acceleration, and there is 0 net force at your foot, so it isn't accelerated.

At the contact patch between your foot and the surface of the earth, there is deformation in response to the force, similar to a spring. In this case it's a "repulsive" force causing two objects to accelerate in opposite directions.

In the case of a car turning on a horizontal surface, the difference in the direction that the front and and rear tires point, and the distance from front to rear tires determines the radius of the turn. Depending on the speed, at the contact patch between tire and road, the tire is pushing outwards on the road, which can only happen if the road is pushing back. Again, it's a "repulsive" force, causing the car to accelerate inwards, and the Earth surface to accelerate outwards (a tiny amount). Since the tires are flexible, they deform significantly at the contact patch, and the actual radius of a turn is greater than the radius calculated by the difference in direction of front and rear tires and the distance between them. At each tire, there is an angle between the direction the tire is pointed, and the actual direction the tire is moving, called the slip angle. As the cornering forces increase, the slip angle increases, until the tire reaches it's limits of traction, in which case the cornering forces start to decrease, radius increases, while slip angle increases (sometimes the cornering forces and radius remain about the same throughout a range of slip angles when near the limits, good for controllability in racing cars).

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tonyh
So here's my point of confusion...where does the force from the ground pushing back on me come from? All I can see here are two forces acting on my foot: 1.) my leg pushing my foot backwards 2.) static friction pushing my foot forwards, resulting in a 0 net force on the foot.

You only need to consider the *external* forces. You are pushing against the floor and the reaction to this is a force on you from the floor. So there is net external force on you which propels you forward.

Can someone please help me understand how the car is able to turn with a physically rigorous description (including forces exerted by the tire, etc.) because I am having the woorrst time trying to understand it!

I understand this as follows: The car is being 'asked' to partake in circular motion (by the turning of the steering wheel), but has a natural tendency to continue its straight line motion (i.e., its inertia). Relative to the desired circular motion, this inertial path is radially outwards and is usually called sideslip. So we have a tendency of sideslip, and in reaction to this the road applies a radially inwards static frictional force to the tires. This provides the required centripetal force. I think this is correct?

Mentor
So this is my understanding thus far. When I wish to walk forward I plant my foot on the floor. Then I use my muscles to apply a force to my foot backwards. If there was no on the surface, the force applied to my foot would cause it to accelerate in a backwards direction and thus cause me to slip. Because there is friction, the force of static friction acts on my foot in the forward direction, balancing the force I applied. This causes my foot to be stationary.

So here's my point of confusion...where does the force from the ground pushing back on me come from? All I can see here are two forces acting on my foot: 1.) my leg pushing my foot backwards 2.) static friction pushing my foot forwards, resulting in a 0 net force on the foot.
Well your foot isn't moving, is it? So the forces are balanced. The force that propels your body forward is not between your foot and the ground, it is between your leg and your foot.

tonyh
Well your foot isn't moving, is it? So the forces are balanced. The force that propels your body forward is not between your foot and the ground, it is between your leg and your foot.
Could you elaborate on this please Russ? Wouldn't this be an internal force? How can an internal force move the centre of mass? There is only one external force on your foot (from the floor). To be balanced, it would require another *external* force wouldn't it?

Mentor
Imagine if you had a spring with a mass on each side sitting horizontally on a smooth surface. You compress the spring with one mass sitting against a wall. When you release the spring, the spring expands and one mass moves while the other does not. The center of gravity also moves. Once the spring is extended a certain amount, it pulls the second mass away from the wall and the entire system is then in motion.
As the spring then compresses again, the second mass catches up to the first one. Think about how a long-jumper does his jump - the second mass (his legs) actually passes the first mass (his torso) in the air.

But perhaps I'm not really addressing the question in the OP:
So here's my point of confusion...where does the force from the ground pushing back on me come from? All I can see here are two forces acting on my foot: 1.) my leg pushing my foot backwards 2.) static friction pushing my foot forwards, resulting in a 0 net force on the foot.
Your leg is like the spring in the example above and it isn't just pushing against your foot, it is pushing against something else, too: your torso. And unless your torso is leaning against a wall, that's the only force pushing against your torso, so it must be balanced by inertia/acceleration.

indoguy427
hmm what I am not understanding here is where static friction is coming into play, both for walking and driving...can someone explain how it comes into play and how in the driving case it provides the centripetal force?

Gold Member
hmm what I am not understanding here is where static friction is coming into play, both for walking and driving...can someone explain how it comes into play and how in the driving case it provides the centripetal force?

1. In Classical physics, forces are exerted by an agent on an object. So, when specifying forces, it is important to state a. what force is being considered, b. what agent is exerting it, and c. on what object it is being exerted. Thus your weight is the gravitational force, exerted BY the gravitational field, ON you.

2. The only forces that matter when considering the motion (or equilibrium, which is a special case of motion) are the forces ON THE BODY OF INTEREST. If any force is not exerted ON the object of interest, it plays no part in determining the motion of that body.

3. When drawing a free-body diagram, ONLY include forces exerted ON the the body being considered. So, for example, if you are interested in how you move, only forces acting on YOU are important. In particular, the reactions to those forces do NOT belong to the free body diagram for you.

4. Forces are exerted in only 2 ways in Classical Mechanics (CM) force fields, and contact. The only force fields you encounter in classical situations are gravitational, and sometimes electomagnetic. The strong and weak nuclear forces play no role in CM.

5. Each contact that the body of interest has with the environment is in turn responsible for two forces: 1. the normal force, exerted BY the surface of contact ON the object of interest, which is always perpendicular to the surface of contact and in the direction that stops the body from moving through the surface, and 2. the friction force, which is always tangential to the surface of contact, in the direction opposite to the motion of the body if the body is sliding, or in the direction opposite to the motion that would occur were the friction absent if the body is not sliding.

So, for a man walking, the only field acting on him is gravitational, acting at his centre of gravity, and the only contact he has is with the ground. The ground exerts a force upwards that stops him dropping into hell, and the friction force exerted by the ground on him. (The force exerted by him on the ground does not belong to his free body diagram, but on that of the Earth - which is of no interest in this problem - since it is a force exerted by the man on the ground, and so is not a force on the man.)

Try to apply these principles to your problems and see if it helps. If not, let me know and I will give you further help.

For circular motion, what you show by the theory is this: if the motion is uniform circular motion, then the TOTAL force exerted ON the particle must be centripetal. (This is not true for non-uniform circular motion.) This does not mean that any actual physical force exerted on the object is necessarily centripetal - it only means that their resultant is centripetal.

Gold Member
hmm what I am not understanding here is where static friction is coming into play, both for walking and driving...can someone explain how it comes into play and how in the driving case it provides the centripetal force?

Rereading the discussion so far, there is another point of confusion that needs to be cleared up: what object are you considering? Before you can apply Newton's laws to determine the motion of an object, you must be clear about which object (or, system) you are considering. In the discussion on "walking", is it the person? Is it the foot? Is it the leg? Is it the torso? The answer you give depends on the system under consideration. So, for example, the answer given by russ_watters "well your foot is not moving, is it" implicitly assumes that you are considering the foot as the system. The foot is not moving relative to the ground. So (assuming that a reference system at rest on the ground is inertial) there has to be zero nett force on it, by N II (or, by N I, for that matter). what forces are exerted on the foot? The friction exerted by the ground on the base of the foot, the normal reaction exerted by the ground on the base of the foot, and the contact force exerted by the leg on the foot. So the sum of these forces must be zero. However, you were not asking about the movement of the foot, but of the person. So to focus on the foot alone as the system is not useful. You need to identify the object of interest first, and then apply the laws to that system.

So, first and most important thing to do is: identify your system. Then apply the method I gave in the previous post.

raja.ganguly
You know what i think guys ! Just guessing--
lets say the person is walking from a viewer's left to his right... then, the contact force b/w foot and ground will have two components: one down right inside to the ground(which is canceled by the Earth's normal reaction); and the other.. to the left side with respect to the viewer. The Earth reciprocates(ofcourse simultaneously, acc. to Newtons third law), and generates a force that has components perpendicular to the ground in the up direction, and another component that points in the forward(or right direction wrt to the viewer) direction.

Ofcourse Earth has negligible acceleration due to the force by foot on it; but the foot will definitely have an definite acceleration forward due to the force by Earth on foot !
So as for the foot to not slip, static friction acts in the back(left direction wrt viewer) direction and balances the forward force... thus stopping the impending motion !

However if i want to deliberately make my foot slip on the ground(obviously with me standing on one leg so that i don't fall ;] ) in the left direction... my foot will be moving left relative to the earth.. so kinetic friction says "hey loser, you don't go that way.. you must come in the right direction"... and thus direction of kinetic friction is in the right direction !

So really, in this case of walking, kinetic and static friction will have opposite directions (with static being towards the viewer's left and kinetic being towards the viewer's right). It really depends on me which kind of friction i want to use.. static or kinetic and depending on that the direction ! If i want to keep my foot stationary relative to the ground, friction will be static and will be towards left; if i don't want it stationary, friction will be kinetic and towards the right !

R Power
I have a general idea that when i put my foot down, I push backwards against the floor and that causes the floor to push me forward. This is a very generic description though and I am trying to decompose the matter into FBD's.

So this is my understanding thus far. When I wish to walk forward I plant my foot on the floor. Then I use my muscles to apply a force to my foot backwards. If there was no on the surface, the force applied to my foot would cause it to accelerate in a backwards direction and thus cause me to slip. Because there is friction, the force of static friction acts on my foot in the forward direction, balancing the force I applied. This causes my foot to be stationary.

So here's my point of confusion...where does the force from the ground pushing back on me come from? All I can see here are two forces acting on my foot: 1.) my leg pushing my foot backwards 2.) static friction pushing my foot forwards, resulting in a 0 net force on the foot.

If someone out there is listening, can you please explain how we are able to walk in terms of individual forces, so I can understand the picture completely?
You are asking something similar to this : you fire a bullet from a gun, gun exerts mechanically some force on bullet, bullet exerts equal and opposite force on gun, so forces are balanced, so why should the bullet accelerate?
I hope you got what I mean to say.
When you make FBD of your foot and Earth system, take one thing at a time.
Heres your solution: Your foot exerts some force on the ground. Ground exerts equal and opposite force on your foot through friction we can say. Now take your foot and make the FBD , you got only one force acting on your foot which is the friction which makes your foot and body accelerate in forward direction.

raja.ganguly
You are asking something similar to this : you fire a bullet from a gun, gun exerts mechanically some force on bullet, bullet exerts equal and opposite force on gun, so forces are balanced, so why should the bullet accelerate?
I hope you got what I mean to say.
When you make FBD of your foot and Earth system, take one thing at a time.
Heres your solution: Your foot exerts some force on the ground. Ground exerts equal and opposite force on your foot through friction we can say. Now take your foot and make the FBD , you got only one force acting on your foot which is the friction which makes your foot and body accelerate in forward direction.

Hey R Power, why should the foot 'accelerate' in the forward direction. I feel, while walking, the foot is stationary relative to the ground, while only the torso and leg move forward !

Your bullet-gun anology has a good point though.. as i too feel that the questioner has made all the forces for the foot !

R Power
Hey R Power, why should the foot 'accelerate' in the forward direction. I feel, while walking, the foot is stationary relative to the ground, while only the torso and leg move forward !
My mistake! the foot transfer the force to the leg and the leg and so the body accelerates.

raja.ganguly
You know what i think guys ! Just guessing--
lets say the person is walking from a viewer's left to his right... then, the contact force b/w foot and ground will have two components: one down right inside to the ground(which is canceled by the Earth's normal reaction); and the other.. to the left side with respect to the viewer. The Earth reciprocates(ofcourse simultaneously, acc. to Newtons third law), and generates a force that has components perpendicular to the ground in the up direction, and another component that points in the forward(or right direction wrt to the viewer) direction.

Ofcourse Earth has negligible acceleration due to the force by foot on it; but the foot will definitely have an definite acceleration forward due to the force by Earth on foot !
So as for the foot to not slip, static friction acts in the back(left direction wrt viewer) direction and balances the forward force... thus stopping the impending motion !

However if i want to deliberately make my foot slip on the ground(obviously with me standing on one leg so that i don't fall ;] ) in the left direction... my foot will be moving left relative to the earth.. so kinetic friction says "hey loser, you don't go that way.. you must come in the right direction"... and thus direction of kinetic friction is in the right direction !

So really, in this case of walking, kinetic and static friction will have opposite directions (with static being towards the viewer's left and kinetic being towards the viewer's right). It really depends on me which kind of friction i want to use.. static or kinetic and depending on that the direction ! If i want to keep my foot stationary relative to the ground, friction will be static and will be towards left; if i don't want it stationary, friction will be kinetic and towards the right !

Hey guys.. can somebody please comment on my idea and tell me if i am right or wrong?? I am sort of confused myself... and there are a lot of different answers i am finding everywhere that has confused my head too !

raja.ganguly
sorry folks.. anybody pls, don't confuse urself with what i wrote. I got my answer ! and what i wrote above is all wrong !

sgstudent
For the walking case, you apply a backwards force on the earth, which can only happen if the surface of the Earth is applying an equal and opposing forward force. (You can't push something unless it pushes back). Since neither you or the Earth are constrained, both you and the Earth are accelerated by this force, but since the Earth is so massive, you experience almost all of the acceleration, in a forwards direction, and the Earth is translated and rotated a very tiny amount in the backwards direction. If you use the Earth as a frame of reference, then you do all of the acceleration, and there is 0 net force at your foot, so it isn't accelerated.

At the contact patch between your foot and the surface of the earth, there is deformation in response to the force, similar to a spring. In this case it's a "repulsive" force causing two objects to accelerate in opposite directions.

In the case of a car turning on a horizontal surface, the difference in the direction that the front and and rear tires point, and the distance from front to rear tires determines the radius of the turn. Depending on the speed, at the contact patch between tire and road, the tire is pushing outwards on the road, which can only happen if the road is pushing back. Again, it's a "repulsive" force, causing the car to accelerate inwards, and the Earth surface to accelerate outwards (a tiny amount). Since the tires are flexible, they deform significantly at the contact patch, and the actual radius of a turn is greater than the radius calculated by the difference in direction of front and rear tires and the distance between them. At each tire, there is an angle between the direction the tire is pointed, and the actual direction the tire is moving, called the slip angle. As the cornering forces increase, the slip angle increases, until the tire reaches it's limits of traction, in which case the cornering forces start to decrease, radius increases, while slip angle increases (sometimes the cornering forces and radius remain about the same throughout a range of slip angles when near the limits, good for controllability in racing cars).

So for the free body diagram would look like this http://www.brighthubeducation.com/s...49-applications-of-newstons-laws-to-walking/#

But how does this relate to the other leg? The other leg would also move forward but in the free body diagram here only shows one leg with a net forward force.