# Direction of frictional force during walking

## Main Question or Discussion Point

Will you please explain where actually the force due to friction acts while a man is walking forward.
I am really confused with a lot of confusing things in the internet.
I am unable to give a fixed answer to my students.

I saw somewhere that friction acts in forward direction, please explain it to me.

Since this is my first post in here, I don't actually know the rules, please consider.

Related Other Physics Topics News on Phys.org
A.T.
Gold Member
Friction always opposes the relative motions between two surfaces in contact.
So when the foot moves from left to right,friction is from right to left

I am really confused with a lot of confusing things in the internet.
What is the website where you got confused?

Zz

Chestermiller
Mentor
When you are walking forward, the bottom of your rear foot exerts a backward frictional force on the ground. The ground exerts an equal and opposite forward force on your rear foot. This force moves you forward.

A.T.
Friction always opposes the relative motions between two surfaces in contact.
So when the foot moves from left to right,friction is from right to left
He is asking about normal walking, not sliding around. In normal walking there is no relative motion between shoe and ground during most of the stance phase. The horizontal ground reaction force is static friction, not dynamic friction. Its direction is the same as the horizontal acceleration of the body's center of mass.

A.T.
When you are walking forward, the bottom of your rear foot exerts a backward frictional force on the ground. The ground exerts an equal and opposite forward force on your rear foot. This force moves you forward.
Only in the late stance phase. In the early stance phase it is the other way around, and you are braking. See diagram from above link:

Chestermiller
Mentor
Hi A.T.:

Very interesting. Am I correct in saying that the integral of the frictional force by the ground over each gait cycle has to be positive? I can't tell from the figure whether this is the case.

Chet

A.T.
Am I correct in saying that the integral of the frictional force by the ground over each gait cycle has to be positive?
If your final horizontal speed equals your initial horizontal speed, the integral is only slightly positive due to air resistance.

AlephZero
Homework Helper
Am I correct in saying that the integral of the frictional force by the ground over each gait cycle has to be positive?
If your final horizontal speed equals your initial horizontal speed, the integral is only slightly positive due to air resistance.
If all the friction is static (i.e. no sliding), the friction force does no work. The work done to overcome air resistance comes from your muscles, not from friction. The work is done in changing the angles of the joints in the legs.

I don't see any obvious reason why the time-average of the friction force should be take any particular value. You could speculate that it the average would be in different directions depending on how you were moving (e.g. sprinting, or goose-stepping)

Chestermiller
Mentor
Thanks AT and AlephZero. Excellent answers.

Chet

Gold Member
I think we should wait for the OP to return now.

A.T.
The work done to overcome air resistance comes from your muscles, not from friction.
Work is not everything. Momentum conservation still applies. At constant average speed the horizontal momentum transfer from the ground, must cancel the horizontal momentum transfer to the air.

You could speculate that it the average would be in different directions depending on how you were moving (e.g. sprinting, or goose-stepping)
Different speeds have different air resistance, and therefore require different horizontal momentum transfer from the ground, in order to maintain a constant average speed.

AlephZero
Homework Helper
I don't see any obvious reason why the time-average of the friction force should be take any particular value.
Thinking a bit more, from the free body diagram for the human, the time-average friction force on the ground + the time average horizontal air resistance force = 0, for constant average walking velocity.

(And that seems like a different way to state A.T.s momentum transfer argument).

A.T.