For the friction formula when to use Fn(normal force)×mu and when F×mu

In summary, the friction formula using Fn (normal force) × μ is applied when calculating the frictional force acting on an object resting on a surface, where μ is the coefficient of friction. Conversely, F × μ is used when dealing with scenarios involving forces other than the normal force, such as when an external force is applied to an object that affects the frictional force. Understanding the context and forces at play is crucial for selecting the appropriate formula.
  • #1
Iamconfused123
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Homework Statement
When to use for the friction formula; Fn(normal force)×mu and when to use F×mu?
Force F is pushing the object against the wall, will the body move or not?
Relevant Equations
F=ma, Ffr=Fn×mu, Ffr=F×mu
Is there some kind of rule when we are supposed to use Fn×mu compared to F×mu or mg×mu to calculate friction?
Because I don't get the same answer when I use Fn×mu and F×mu. In magnitude the forces are the same but they are not same in directions.
Why is it so?

Here is the photo. Problem goes; will the body(0.5 kg) move when pressed with force F=12 N against wall with static coef. of friction = 0.6.
It won't, but that is not what bothers me.

CamScanner 2023-12-27 00.56_1.jpg


Like I know that for case no.2 I am supposed to subtract, but I wonder why I didn't end up with one force as negative. If we pretend that I don't know that I am supposed to subtract one force from another, how would I end up with -2.2 and not 12.2? Because subtraction is additon of negative number and I don't have any negatives in case no.2.

I know that it's because F and Fn are going in opposite directions and have different signs, but even in most basic problems when calculating Friction force we can use Ffr= Fn×mu or Ffr=mg×mu. And those answers were always the same, or rather I figured that I was subtracting two positives using my knowledge that forces act in opposite direction so it must be subtraction, like, I came to conclusion using intelligence, it didn't conclude itself from math.
So, why is then in textbooks written that Ffr=mg×mu, when in fact is Ffr= Fn × mu.? In magnitude it's the same as Ffr=mg×mu but not in the dirrection.
 

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  • #2
It is always the normal component of the net force acting on the object.
 
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  • #3
nasu said:
It is always the normal component of the net force acting on the object.
Noting of course that this is the maximal friction. Friction in a static situation will typically not be equal to normal force x friction coefficient, but rather smaller than that - unless it is really at the limit of slipping.
 
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  • #4
nasu said:
It is always the normal component of the net force acting on the object.
Oh, didn't know that. Thank you.

But why do people write Fn=mg, when it's not. It's Fn=-mg

Like this for example.
pf2.png

Why didn't he write Fn=-mg insted of Fn=mg. It's the same in my textbook, and I believe I have seen the same expressions elsewhere.
 
  • #5
Orodruin said:
Noting of course that this is the maximal friction. Friction in a static situation will typically not be equal to normal force x friction coefficient, but rather smaller than that - unless it is really at the limit of slipping.
Thanks, I know. But I really appreciate the comment.
 
  • #6
Iamconfused123 said:
Oh, didn't know that. Thank you.

But why do people write Fn=mg, when it's not. It's Fn=-mg

Like this for example.View attachment 337744
Why didn't he write Fn=-mg insted of Fn=mg. It's the same in my textbook, and I believe I have seen the same expressions elsewhere.
The positive directions of the forces have been specified in the figure. In opposite directions.
 
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  • #7
Orodruin said:
The positive directions of the forces have been specified in the figure. In opposite directions.
Ohh, so if we draw a sketch we can just do the algebra?
 
  • #8
The
Iamconfused123 said:
Oh, didn't know that. Thank you.

But why do people write Fn=mg, when it's not. It's Fn=-mg

Like this for example.View attachment 337744
Why didn't he write Fn=-mg insted of Fn=mg. It's the same in my textbook, and I believe I have seen the same expressions elsewhere.
The sign of the force component depends on the choice of positive direction. The sign is not relevant for the formula for friction force (static or kinetic) which relates the magnitudes of the two forces (normal and friction).
 
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  • #9
I should add: It is quite common to introduce force directions in different directions depending on what you would expect to be the force direction. There is of course nothing wrong with doing so as long as you take the correct sign in the equilibrium equation. If your intuition about the direction was wrong, then the force will simply cone out negative, meaning it is in the opposite direction of what you expected.

Of course, there is also nothing wrong with defining all forces with the same positive direction. The sign will then tell you the actual direction.
 
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  • #10
Orodruin said:
I should add: It is quite common to introduce force directions in different directions depending on what you would expect to be the force direction. There is of course nothing wrong with doing so as long as you take the correct sign in the equilibrium equation. If your intuition about the direction was wrong, then the force will simply cone out negative, meaning it is in the opposite direction of what you expected.

Of course, there is also nothing wrong with defining all forces with the same positive direction. The sign will then tell you the actual direction.
Thanks, but just to confirm. If we draw a sketch we can just do the algebra like when we are doing normal math equations?

Because it kind of does not make sense that Fn=mg. If he wrote |Fn|=|mg| then okay. But forces are vectors.
 
  • #11
nasu said:
The
The sign of the force component depends on the choice of positive direction. The sign is not relevant for the formula for friction force (static or kinetic) which relates the magnitudes of the two forces (normal and friction).
But shouldn't we then use absolute values for that? |Fn|=|mg|
 
  • #12
Iamconfused123 said:
If we draw a sketch we can just do the algebra like when we are doing normal math equations?
As long as you take the appropriate signs into account when writing down the equilibrium equation. In your example, weight is defined in the down direction and the normal force in the up direction. The equilibrium equation in the up direction is therefore Fn - W = 0 (not Fn + W = 0) because weight was defined positive in the opposite direction of the direction you are considering. Therefore Fn = W. Since the weight is mg downwards and weight was specified as positive downwards, Fn = mg. Since Fn was specified upwards, Fn is indeed equal to mg upwards.
 
  • #13
nasu said:
It is always the normal component of the net force acting on the object.
… do you mean, the normal component of the net of the other forces acting on the object? But that would still be only for statics. For an accelerating body the normal force might not equal that.
I define the normal force, ##F_n##, as the minimum reaction force from the contacting body to prevent interpenetration.
Necessarily, the minimum such force will be normal to the contact plane between the bodies. (For corner-on-corner contact the contact plane is indeterminate.)
 
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  • #14
haruspex said:
Yes. The correct forms are
Static friction: |Fs|≤μs|Fn|
Kinetic friction: |Fk|=μk|Fn|
where the normal force, Fn, is the minimum reaction force from the contacting body to prevent interpenetration.
This was not the question. The question was if it should be ##|F_n| = |mg|## or not. This equation, while true, removes the information about the proper direction of the normal force.
 
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  • #15
haruspex said:
… to be more precise, the normal component of the net of the other forces acting on the object. Which leads to another potentially indeterminate case, where the object contacts multiple other bodies.
Other than what?
But yes, for the case of multiple contacts I would say just the normal component of the contact force. It is not the net force.
But thus is going too far from the OP question. Which is the usual way here. 😊
 
  • #16
Orodruin said:
This was not the question. The question was if it should be ##|F_n| = |mg|## or not. This equation, while true, removes the information about the proper direction of the normal force.
Whoops. Thanks, corrected.
 
  • #17
nasu said:
Other than what?
But yes, for the case of multiple contacts I would say just the normal component of the contact force. It is not the net force.
But thus is going too far from the OP question. Which is the usual way here. 😊
If the body is not accelerating, the net force on it is zero, so post #2 cannot be what you meant.
 
  • #18
Oh, other than the force the surface (other object) acts on the body. Yeah, thank you for clarification. I just wondered what you mean. I was not careful in my post.
 

FAQ: For the friction formula when to use Fn(normal force)×mu and when F×mu

When should I use the friction formula Fn × μ?

You should use the friction formula Fn × μ when you are calculating the frictional force between two surfaces in contact. Here, Fn represents the normal force, which is the perpendicular force exerted by the surface on the object, and μ is the coefficient of friction. This formula is typically used to determine the maximum static friction or kinetic friction.

What does Fn represent in the friction formula?

In the friction formula, Fn represents the normal force. This is the force exerted by a surface perpendicular to the object resting on it. For example, if an object is placed on a horizontal surface, the normal force is equal to the weight of the object. If the surface is inclined, the normal force is the component of the object's weight perpendicular to the surface.

When do I use F × μ instead of Fn × μ?

Generally, you should use Fn × μ rather than F × μ. The formula F × μ is not standard and might be a misunderstanding. The correct formula for frictional force involves the normal force (Fn) and the coefficient of friction (μ). Ensure you are using the normal force in your calculations to avoid errors.

What is the coefficient of friction (μ) and how do I determine it?

The coefficient of friction (μ) is a dimensionless scalar value that represents the frictional properties of the surfaces in contact. It varies depending on the materials and surface conditions. You can determine μ experimentally by measuring the forces involved or look it up in tables that provide typical values for different material pairs.

How does the angle of the surface affect the normal force (Fn) in the friction formula?

The angle of the surface affects the normal force because it changes the component of the gravitational force acting perpendicular to the surface. On an inclined plane, the normal force is calculated as Fn = mg cos(θ), where m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of inclination. This reduced normal force affects the frictional force calculated using Fn × μ.

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