Friction force on each foot and on each hand of a person

In summary: I'll post the equations below.In summary, Mentor explains how to solve for the normal force acting on a person's feet and hands using Newton's First Law. However, the normal force cannot be solved for when there is friction between the person's skin and the ice. If the ground were frictionless, the person would slip outwards.
  • #1
dl447342
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Homework Statement
The yoga exercise “Downward- Facing Dog” requires stretching your hands straight out above your head and bending down to lean against the floor. This exercise is performed by a 750 N person as shown in Fig. P11.67. When he bends his body at the hip to a 90° angle between his legs and trunk, his legs, trunk, head, and arms have the dimensions indicated. Furthermore, his legs and feet weigh a total of 277 N, and their center of mass is 41 cm from his hip, measured along his legs. The person’s trunk, head, and arms weigh 473 N, and their center of gravity is 65 cm from his hip, measured along the upper body.

(a) Find the normal force that the floor exerts on each foot and on each hand, assuming that the person does not favor either hand or either foot.

(b) Find the friction force on each foot and on each hand, assuming that it is the same on both feet and on both hands (but not necessarily the same on the feet as on the hands).
Relevant Equations
Equation for net torque: sum of ##\vec{r}\times \vec{F}## for all forces ##\vec{F}## and their position vectors.

Newton's first law; if an object is not accelerating, the net force on it must be zero.
I get how to solve (a); my method involves finding the net torque about the man's hands and setting it to zero, which can be used to solve for the normal force acting on his feet and the normal force on his hands can be solved using Newton's first law. Then divide by 2 for each to get the normal force on one hand or foot.

However, I'm not sure how to get part (b); neither setting the net torque about any point to 0 nor setting the net force to zero works because it seems the friction forces on his hands and feet are equal in magnitude and opposite in direction, so their torques about any point would always sum to zero.

[Image added by Mentors]
Downward-Facing-Dog-Pose-Adho-Mukha-Svanasana.jpg

https://www.ekhartyoga.com/media/im...ward-Facing-Dog-Pose-Adho-Mukha-Svanasana.jpg
 
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  • #2
dl447342 said:
However, I'm not sure how to get part (b); neither setting the net torque about any point to 0 nor setting the net force to zero works because it seems the friction forces on his hands and feet are equal in magnitude and opposite in direction, so their torques about any point would always sum to zero.
What would happen if you tried to do down dog on frictionless ice?
 
  • #3
dl447342 said:
neither setting the net torque about any point to 0 nor setting the net force to zero works because it seems the friction forces on his hands and feet are equal in magnitude and opposite in direction, so their torques about any point would always sum to zero.
The question presumably intends, but omits to state, that you should treat the hips as a free joint, with no muscular tension to maintain the angle at 90°.
 
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  • #4
PeroK said:
What would happen if you tried to do down dog on frictionless ice?
@PeroK thanks for the hint but can you provide some equations?

From my understanding, one should find the component of the normal force along his legs and find the component of this force along the ground. This gives ##F_s = N\cos \theta \cos \phi## where ##\theta## is the angle between the normal force on his legs and his legs and ##\phi## is the angle between his legs and the ground.

But why can't one apply this method to his hands?

Should friction point outwards or inwards in this case? I think intuitively if the ground were frictionless, the person would slip outwards and since friction opposes motion I think friction should point inwards.
 
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  • #5
dl447342 said:
@PeroK thanks for the hint but can you provide some equations?

From my understanding, one should find the component of the normal force along his legs and find the component of this force along the ground. This gives ##F_s = N\cos \theta \cos \phi## where ##\theta## is the angle between the normal force on his legs and his legs and ##\phi## is the angle between his legs and the ground.

But why can't one apply this method to his hands?

Should friction point outwards or inwards in this case? I think intuitively if the ground were frictionless, the person would slip outwards and since friction opposes motion I think friction should point inwards.
As I posted, you need to treat the hips as a free joint. Consider torque balance on the portion to one side of that.
 
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  • #6
Note that there is a maximum friction force for hands and for feet, which is proportional to the coefficient of friction between skin and floor and to N force on hands and on feet.
The person may or may not need to reach those magnitudes of friction for keeping that position, only the friction forces that are sufficient to avoid a skid in one or both directions.

Yoga.png
 
  • #7
dl447342 said:
@PeroK thanks for the hint but can you provide some equations?
Down dog + ice = flat on your face
 
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  • #8
haruspex said:
As I posted, you need to treat the hips as a free joint. Consider torque balance on the portion to one side of that.
Thanks. I figured it out.
 

Related to Friction force on each foot and on each hand of a person

1. What is friction force?

Friction force is the force that opposes the motion of an object when it comes into contact with another surface. It is caused by the microscopic irregularities on the surface of the two objects rubbing against each other.

2. How is friction force calculated?

Friction force is calculated by multiplying the coefficient of friction (a measure of how rough or smooth the surfaces are) by the normal force (the force perpendicular to the surface). This gives us the maximum possible friction force, but the actual force may be lower depending on the situation.

3. How does friction force affect a person's movement?

Friction force on each foot and hand of a person can affect their movement by providing the necessary grip and traction to walk, run, or hold onto objects. It also helps to maintain balance and prevent slipping or falling.

4. How does the friction force on each foot and hand differ?

The friction force on each foot and hand can differ depending on the surface they are in contact with, the type of footwear or gloves being worn, and the amount of force being applied. The friction force on hands is usually higher due to their smaller surface area and the ability to apply more pressure.

5. How can friction force be reduced?

Friction force can be reduced by using lubricants, such as oil or grease, between two surfaces. It can also be reduced by using smoother surfaces or by decreasing the force applied. In some cases, adding weight to increase the normal force can also decrease friction force.

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