Normal Forces

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ln(
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I thought I understood normal forces but I am confused. Could someone explain them to me?

What is the normal force on you if you are hanging or climbing up a vertical rope? In a definition, it says that the normal force is perpendicular to the surface, but if the surface is the rope, then the normal force is horizontal...?
 

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  • #2
Baluncore
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Yes, horizontal.
The term “normal” is being used here to mean orthogonal or “at right angles to”.
 
  • #3
ln(
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Yes, horizontal.
The term “normal” is being used here to mean orthogonal or “at right angles to”.
But on a rope, you are applying your weight force downwards, not towards the rope. So it doesn't make sense that the normal force is horizontal, as in away from the rope. When climbing, you want the normal force to be upwards.
 
  • #4
ln(
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Ahh, never mind I understand now.

The normal force of the rope when hanging down is horizontal, so if you apply a squeezing force in a horizontal fashion, you are able to oppose your own weight.
 
  • #5
ehild
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The normal force is horizontal, but the force opposite to your weight is the force of friction, which is parallel to the rope. If you do not move on the rope, it is static friction, which maximum value is proportional to the normal force.

As you grab the rope and squeeze it, you exert force round it, and the rope exerts normal force to your whole palm outward. The net force on your palm is zero.

ehild
 
  • #6
HallsofIvy
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But on a rope, you are applying your weight force downwards, not towards the rope. So it doesn't make sense that the normal force is horizontal, as in away from the rope. When climbing, you want the normal force to be upwards.
So all you are saying is that there is NO "normal" force in this situation. No, you do NOT "want the normal force to be upward, you want the resuiltant force to be upward in order to climb a rope.
 
  • #7
Pythagorean
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Ahh, never mind I understand now.

The normal force of the rope when hanging down is horizontal, so if you apply a squeezing force in a horizontal fashion, you are able to oppose your own weight.
You got it! Basically, the force that increases the friction force is the normal force, because it increases the intensity of the contact between the two surfaces.
 
  • #8
ln(
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Alright I got it now. You want to have a great enough normal force, as in force you squeeze around the rope, to increase the frictional force and be able to climb the rope.

So in a sense, is the normal force similar to Newton's Third Law?

Also, in the rope climbing situation, how do I calculate how much normal force, and therefore frictional force when multiplied by the frictional constant, is required to accelerate a certain amount using the equation N = mg/u when climbing at a constant velocity or simply hanging there?
Do I rearrange the equation to be uN = mg, and subtract mg from uN, and can find the net force that can be able to be utilized to accelerate?
 
  • #9
Pythagorean
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Normal force wouldn't be possible without N3, so it's connected.

uN = mg comes from ma = 0 (no acceleration)

you start with F = uN-mg

N2 gives:

ma = uN-mg

for a=0, this becomes uN=mg.

now what if a is not 0?
 
  • #10
ln(
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Normal force wouldn't be possible without N3, so it's connected.

uN = mg comes from ma = 0 (no acceleration)

you start with F = uN-mg

N2 gives:

ma = uN-mg

for a=0, this becomes uN=mg.

now what if a is not 0?
I would guess that if a is <0, then you will start to fall due to gravity, but if it is >0, you have the friction needed to accelerate, but it does not mean that necessarily you are accelerating. Simply because you have more friction than needed to oppose the force of gravity does not mean you are accelerating, you could simply be grabbing the rope with more force than needed...

However the equation says otherwise. It says that if the net force is over 0, you are accelerating. Which may or may not be true.
 
  • #11
Pythagorean
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just to find what N is required for a given a, you take:

ma = uN-mg

and solve for N, assuming some constant acceleration, a, you wish to achieve, just as you did when a=0, you just don't lose the 'ma' term this time since a is not 0. Of course, once you're in motion, you only need to maintain velocity, so you can lower the normal force (your grip) back to the a=0 case. In reality, the act of climbing would probably involve a lot of stopping and starting with each hand-over, though.
 

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