# Doubts about the mechanics of walking

• Freaky Fred

#### Freaky Fred

Hello
When we are walking, exist, basically, three forces on us foot.(gravity, normal and friction forces).
Considering that the frictional force is in static regime, the resulting force is 0. Theoretically,
there is no relative movement between the foot and the floor. So my doubts:
How the body is projected to front and how the foot position has influence about that ?

Considering that the frictional force is in static regime, the resulting force is 0.
That is not correct. The friction force is what it is, and it is unopposed, creating a positive resultant force.

Freaky Fred
That is not correct. The friction force is what it is, and it is unopposed, creating a positive resultant force.
Oh sorry. A terrible mistake. But I can not understand why the foot does not slip if the resulting horizontal force is different from 0. I know that in a static regime there is no movement between the surfaces, but I can not understand how this happens if the result is not 0 on the foot.

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Let's see your free body diagram of the foot.

SammyS, CWatters, Freaky Fred and 1 other person
Oh sorry. A terrible mistake. But I can not understand why the foot does not slip if the resulting horizontal force is different from 0. I know that in a static regime there is no movement between the surfaces, but I can not understand how this happens if the result is not 0 on the foot.

What Russ means is that the force of friction is not zero. There are no other frictional forces that somehow oppose the frictional force on your foot. What makes the net force on your foot zero is the force applied to the foot from the ankle and leg.

Freaky Fred and russ_watters
But I can not understand why the foot does not slip if the resulting horizontal force is different from 0. I know that in a static regime there is no movement between the surfaces, but I can not understand how this happens if the result is not 0 on the foot.
To expand on what the others said, this is a different question than the first one you asked. The broader issue is one of correctly/incorrectly selecting or worse mixing together separate models of a scenario in order to answer certain questions to a certain level of accuracy.

Your first question modeled the human body as a single, solid object, with three forces applied to it. The answer you got for this model works for explaining why the body moves, but it is very simplistic.

The second question models the body as more than one part (total number not specified), correctly noting that the plant foot is in fact stationary. So it has four forces acting on it, not three. And if your foot and torso are separate objects, then friction isn't what makes your torso move. So how complicated do you want to make this model...?

Freaky Fred
To expand on what the others said, this is a different question than the first one you asked. The broader issue is one of correctly/incorrectly selecting or worse mixing together separate models of a scenario in order to answer certain questions to a certain level of accuracy.

Your first question modeled the human body as a single, solid object, with three forces applied to it. The answer you got for this model works for explaining why the body moves, but it is very simplistic.

The second question models the body as more than one part (total number not specified), correctly noting that the plant foot is in fact stationary. So it has four forces acting on it, not three. And if your foot and torso are separate objects, then friction isn't what makes your torso move. So how complicated do you want to make this model...?

I really mixed up two ways of interpreting (sorry, again). What I really want to know is how this works considering the body as several parts (foot, leg, trunk, neck, head). I imagine that the force of the leg on the foot will be equal in intensity to the frictional force (so that the foot remains stationary). The leg will receive a force applied by the torso and another force of intensity equal to that applied to the foot. I imagine this will occur with the other parts of the body as well. In these parts of the body (except the foot), the resulting force can not be 0. What are the relationships between forces for the body to be projected forward?

If I consider infinite stacked horizontal planes that mark each part of the body and considering these interceptions of the same mass, can I say that the intensity of the horizontal forces will decrease as the distance they will be from the foot? For example, the part of the body that is moving and closest to the foot (standing parts) is receiving a force from the standing part and a force from the next part above. The force coming from the part that is without movement must have a greater intensity than the opposing force applied by the upper part. So this slice will move forward. The upper part in question will receive a force of the same intensity as it applies in the lower part (the '' first '' moving part), but for it to also have movement, the part above the 'second' part in motion it needs to apply a force of even less intensity than the force that this second part receives from the first. Following this logic, each slice above will receive forces of lower intensities, but with a resultant force equal to each piece.

Thank you guys who are helping me to understand this situation and sorry for my limited English.

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I really mixed up two ways of interpreting (sorry, again). What I really want to know is how this works considering the body as several parts (foot, leg, trunk, neck, head). I imagine that the force of the leg on the foot will be equal in intensity to the frictional force (so that the foot remains stationary). The leg will receive a force applied by the torso and another force of intensity equal to that applied to the foot. I imagine this will occur with the other parts of the body as well. In these parts of the body (except the foot), the resulting force can not be 0. What are the relationships between forces for the body to be projected forward?

You need to be more careful when you say 'the resulting force'. The net force on any part of the body is zero unless that part is accelerating. The individual forces are not zero however.

If I consider infinite stacked horizontal planes that mark each part of the body and considering these interceptions of the same mass, can I say that the intensity of the horizontal forces will decrease as the distance they will be from the foot? For example, the part of the body that is moving and closest to the foot (standing parts) is receiving a force from the standing part and a force from the next part above. The force coming from the part that is without movement must have a greater intensity than the opposing force applied by the upper part. So this slice will move forward. The upper part in question will receive a force of the same intensity as it applies in the lower part (the '' first '' moving part), but for it to also have movement, the part above the 'second' part in motion it needs to apply a force of even less intensity than the force that this second part receives from the first. Following this logic, each slice above will receive forces of lower intensities.

In general the forces weaken as you move upwards, primarily because you have less weight to hold up against gravity. A carefully drawn force diagram (free body diagram) using a few simple 'bones' should clearly show which forces are acting where and how. By 'bones' I mean that you can represent large sections of the body with a single bone. For example, the entire leg could be represented by a single bone. How complicated you make this diagram is up to you.

russ_watters and Freaky Fred
The leg will receive a force applied by the torso and another force of intensity equal to that applied to the foot.
What makes you think the force of the foot on the leg and the force of the torso on the leg are equal and opposite? The leg is accelerating forward.

What makes you think the force of the foot on the leg and the force of the torso on the leg are equal and opposite? The leg is accelerating forward.
That was not what I meant. I wanted to write that the force that the leg applies to the foot have an intensity equal to the force that the foot applies to the leg. The force of the torso on the leg would be of lesser intensity, so that was the first forward movement. As I said in the second part, I imagine that the parts of the body more distant from the foot will receive less intense forces. So my doubt about this point is the relationship between these forces. (Example: if the forces even with different intensities will result in the same force resulting in each point or if the resulting forces will be different for each point (generating different accelerations) at a certain instant).

Intuitively, it looks like the head has the same torso speed, but this is probably a situation when the resulting force is already 0 for those parts.

In general the forces weaken as you move upwards, primarily because you have less weight to hold up against gravity.
Supporting the weight of the above segments is the smaller part of the forces at the joints. The bigger part are muscle forces. For example at the knee and hip you have 3-4 x your bodyweigth during normal walking.

Freaky Fred
I imagine that the parts of the body more distant from the foot will receive less intense forces.
Your slice model applies to a static body, that is accelerated horizontally as whole by a force at the bottom. But human bodies with moving segments are more complex.

Freaky Fred
Your slice model applies to a static body, that is accelerated horizontally as whole by a force at the bottom. But human bodies with moving segments are more complex.
Really. Nature is a beautiful and complex puzzle.

Really. Nature is a beautiful and complex puzzle.
Well, it apllies to robots and vehicles with moving parts too.

I want to throw in another question to this topic:

Why do people claim that there is no work done when carrying a bag horizontally? With respect to the gravitational force, it is clear.
But we still apply a force to push the bag forward. With the movement of our legs/foot we push against the Earth and, through actio=reactio, the Earth pushes us forward and we move. In other words, it's like applying a force to a body of our mass + the mass of the bag, pushing it forward. There should be work involved here.

In other words, it's like applying a force to a body of our mass + the mass of the bag, pushing it forward. There should be work involved here.
If you were constantly pushing the bag forward then you would expect the bag to accelerate to a speed very much higher than a walking pace. The net forward force on a bag of groceries being carried to the car is roughly equal to the wind resistance said bag experiences during the trip: negligible.

CWatters
I want to throw in another question to this topic:

Why do people claim that there is no work done when carrying a bag horizontally? With respect to the gravitational force, it is clear.
But we still apply a force to push the bag forward. With the movement of our legs/foot we push against the Earth and, through actio=reactio, the Earth pushes us forward and we move. In other words, it's like applying a force to a body of our mass + the mass of the bag, pushing it forward. There should be work involved here.
It depends on the exact wording of the question. If you are asking about work done on the bag, the answer is zero. If you are asking about the energy expended by the person, the answer is non-zero.

Humans are inefficient machines. However, note that we learn or evolved to be as efficient as we can be by smoothing the walking stride both horizontally and vertically.

"Why is carrying a bag and walking not considered to be work" . . . There's some relevant info that can be added for better clarity. . . . When a trolley with stuff is pushed, work is said to be done as the force is in the same direction as the displacement. . . . but. . . . . When carrying a bag and walking, the force is in the vertically upward direction, against gravity, while the displacement is in the horizontal direction . . . Since the force and displacement is perpendicular to each other and NOT in the same direction, it is concluded that no work is done.

With an ideal trolley that starts and ends at rest, runs on level ground, and experiences no friction, no work is done.

If you were constantly pushing the bag forward then you would expect the bag to accelerate to a speed very much higher than a walking pace. The net forward force on a bag of groceries being carried to the car is roughly equal to the wind resistance said bag experiences during the trip: negligible.
So the average over time force my feet are exerting on the ground (in order to push myself forward) just compensates the negligible air resistance?

With an ideal trolley that starts and ends at rest, runs on level ground, and experiences no friction, no work is done.
I was about to say that a trolley is far from ideal. But I just realize you brought that point up as a better analogy to the carried bag who does not even touch the ground.

So the average over time force my feet are exerting on the ground (in order to push myself forward) just compensates the negligible air resistance?
Yes.

SchroedingersLion
Unless it is an automatic trolley u would still be applying a force, for it to move from point A to point B. When force and displacement are both there, even in the absence of friction, work has to be done.

Unless it is an automatic trolley u would still be applying a force, for it to move from point A to point B. When force and displacement are both there, even in the absence of friction, work has to be done.
You have to apply a force to start it, doing positive work. You have to apply a force to stop it, doing negative work. Net zero.

Why would u want to stop it ?. . We're just trying to calculate the work done by a body moving from point A thro point B. .

Why would u want to stop it ?. . We're just trying to calculate the work done by a body moving from point A thro point B. .
So it doesn't destroy the train station and kill all its passengers?

Kidding(?) aside, in hypothetical and real life questions like this you nearly always assume travel from point A to point B involves starting from stationary and coming to a dead stop.

Please note, actual trains often have regenerative braking and can recover some of the energy used to accelerate.

A quick calc tells me a train is on the order of 200x more fuel efficient than a person.

Why would u want to stop it ?. . We're just trying to calculate the work done by a body moving from point A thro point B. .
Because if you do not return it to its initial state of motion, you are concerned with apples but are measuring oranges.

Let me try to be more direct about a possible misunderstanding:
Unless it is an automatic trolley u would still be applying a force, for it to move from point A to point B. When force and displacement are both there, even in the absence of friction, work has to be done.
In the absence of friction, there is no force during constant speed motion, so no work required to move from point A to point B if you recover the work expended to accelerate.

jbriggs444