Why do skin burns occur when friction is applied to the skin?

[ln(]

Today, I was thinking about skin burns after climbing some poles in the park and was wondering this:

I understand that a frictional force is determined with the equation F = μN = μmg.

However, it is also to my understanding that when you drag something on your body across a surface, let's say a rope, you will feel a frictional force opposing your motion. But when you quick swipe your hand or finger across a rope, you will feel a greater frictional force despite no change in the frictional constant and of course the weight of your finger or hand.

Why is this?


Also, the what happens when you yourself add a downwards force, directly influencing the normal force and adding to it? Does the frictional force equation change to F = μ(N+f_app)?

Thank you all!

 

[PhanthomJay]

Today, I was thinking about skin burns after climbing some poles in the park and was wondering this:

I understand that a frictional force is determined with the equation F = μN = μmg.

the Normal force N and your weight mg are not always the same. If you climb the pole which has u = 0.5, at constant speed, then the friction force, F, must be mg, from Newton 1, and thus the normal force, per N = F/u, must be 2mg. If the friction coefficient was 0.1, then N = 10mg, which you most probably cannot exert, so you wouldn't be able to climb the thing.

However, it is also to my understanding that when you drag something on your body across a surface, let's say a rope, you will feel a frictional force opposing your motion.

yes, and the amount of friction force depends upon how much pressure (i.e., force) the rope is appying normal to you.

But when you quick swipe your hand or finger across a rope, you will feel a greater frictional force despite no change in the frictional constant and of course the weight of your finger or hand.

, not necessarily, unless you are applying a greter force in so doing. And again , it is not the weight of your hand or finger that matters, it is the applied normal force that determines the friction force. You could swipe it lightly against your finger or strongly, your choice. The stronger, the more friction, the greater the burn.

Also, the what happens when you yourself add a downwards force, directly influencing the normal force and adding to it? Does the frictional force equation change to F = μ(N+f_app)?

Thank you all!

F = uN, where N is the applied force perprndicular to the rope.

 

[sophiecentaur]

There is also the time factor involved. The heating effect will also depend upon how quickly the energy is dissipated by friction. A higher temperature will be reached if the same amount of energy is dissipated in a shorter time because the flow of heat away will be governed by the thermal conductivity of skin etc. and the temperature gradient.
The Power dissipated will be equal to Force times 'Sliding' speed.

 

[ln(]

the Normal force N and your weight mg are not always the same. If you climb the pole which has u = 0.5, at constant speed, then the friction force, F, must be mg, from Newton 1, and thus the normal force, per N = F/u, must be 2mg. If the friction coefficient was 0.1, then N = 10mg, which you most probably cannot exert, so you wouldn't be able to climb the thing.yes, and the amount of friction force depends upon how much pressure (i.e., force) the rope is appying normal to you. , not necessarily, unless you are applying a greter force in so doing. And again , it is not the weight of your hand or finger that matters, it is the applied normal force that determines the friction force. You could swipe it lightly against your finger or strongly, your choice. The stronger, the more friction, the greater the burn.
F = uN, where N is the applied force perprndicular to the rope.

Ahh okay.

So this means that the normal force is the force you yourself must exert to climb at a constant speed? Exerting what the normal force is will allow you to climb 1 m/s? Or simply allow you to not accelerate towards the ground? How do I calculate this? If force applied is greater than normal force, force_app - normal force, equals ma, and I can find acceleration?

Probably not that easy :).

 

[jbriggs444]

So this means that the normal force is the force you yourself must exert to climb at a constant speed? Exerting what the normal force is will allow you to climb 1 m/s? Or simply allow you to not accelerate towards the ground? How do I calculate this? If force applied is greater than normal force, force_app - normal force, equals ma, and I can find acceleration?

Probably not that easy :).

http://en.wikipedia.org/wiki/Normal_force

"normal" in this context means "perpendicular to". If you are standing on level ground, your "normal force" on the ground is equal to your weight. If you are clinging to a vertical rope the "normal force" is how hard you are squeezing.

By contrast, a "tangential" force would be parallel to the surface.

http://en.wikipedia.org/wiki/Normal_force

 

[PhanthomJay]

Oversimplifying the complexity of a human machine, if you hang on to the rope or climb it at any constant speed, the friction force is uN, where N is the 'squeezing' force and uN is mg, per Newton 1 in the vertical direction. Thus, N = mg/u. Doesn't matter waht speed, as long as it is consatnt.
Now if you were to accelerate up the rope, the friction force up must be greater than your weight down, per Newton 2 in the vertical direction, and thus, you must squeeze even harder than mg/u to accelerate, or else you'll start to fall dowm. So if you try to climb a rope fast, you first must accelerate up to that speed, which is difficult because of the large squeeze force required. If you somehow manage to treach thst speed and then climb the rest of the way at that speed, constant sppeed, then the squeeze force is back to mg/u again.

 

[ln(]

Oversimplifying the complexity of a human machine, if you hang on to the rope or climb it at any constant speed, the friction force is uN, where N is the 'squeezing' force and uN is mg, per Newton 1 in the vertical direction. Thus, N = mg/u. Doesn't matter waht speed, as long as it is consatnt.
Now if you were to accelerate up the rope, the friction force up must be greater than your weight down, per Newton 2 in the vertical direction, and thus, you must squeeze even harder than mg/u to accelerate, or else you'll start to fall dowm. So if you try to climb a rope fast, you first must accelerate up to that speed, which is difficult because of the large squeeze force required. If you somehow manage to treach thst speed and then climb the rest of the way at that speed, constant sppeed, then the squeeze force is back to mg/u again.

Okay. I understand.

Just wondering, what are you referring to when you say "per Newton 1" or "per Newton 2"?

 

[sophiecentaur]

Friction, on its own is hardly likely to produce a burn (i.e. heating effect). You need slipping to dissipate energy- then the Power is force times velocity. Isn't this what the OP question was about? The situation in that case is more complicated as there are many more combinations of variables. You can burn your hands without even leaving the ground if you grip lightly and move them downwards fast.
It's speed of slipping and not speed of climbing that counts here.

 

[PhanthomJay]

Okay. I understand.

Just wondering, what are you referring to when you say "per Newton 1" or "per Newton 2"?

Per Newton 1 I mean "using Newton's first law" where the weight force down must be equal to the friction force up(parallel to pole). Using his second law for the accelerating case, the friction force must be greater than the weight, in order that there be a net force upward to accelerate you upward. This is in response to your questions regarding speed, acceleration, and value of friction force. Regarding friction burn, see Sophie's responses.

 

[ln(]

Alright I understand. Thanks everyone!

Time to study the heating effect...

 

[ln(]

Oversimplifying the complexity of a human machine, if you hang on to the rope or climb it at any constant speed, the friction force is uN, where N is the 'squeezing' force and uN is mg, per Newton 1 in the vertical direction. Thus, N = mg/u. Doesn't matter waht speed, as long as it is consatnt.
Now if you were to accelerate up the rope, the friction force up must be greater than your weight down, per Newton 2 in the vertical direction, and thus, you must squeeze even harder than mg/u to accelerate, or else you'll start to fall dowm. So if you try to climb a rope fast, you first must accelerate up to that speed, which is difficult because of the large squeeze force required. If you somehow manage to treach thst speed and then climb the rest of the way at that speed, constant sppeed, then the squeeze force is back to mg/u again.

Upon reading this again, I do not understand N = mg / u. Although I understand the equation, shouldn't N be equal to the force you push downwards, not squeezing force? Because in order to climb, you need to apply more force downwards to oppose your weight and apply a net force if you want to accelerate...?