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

[dl447342]

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]

https://www.ekhartyoga.com/media/im...ward-Facing-Dog-Pose-Adho-Mukha-Svanasana.jpg

 

[PeroK]

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?

 

[haruspex]

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°.

 

[dl447342]

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.

 

[haruspex]

@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.

 

[Lnewqban]

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.

 

[PeroK]

@PeroK thanks for the hint but can you provide some equations?

Down dog + ice = flat on your face

 

[dl447342]

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.