Contact and electromagnetic force

[rudransh verma]

Homework Statement

1. F, Fn, f are contact force, normal force, friction exerted by one surface over another. If none is zero then-
A. F>Fn
B. F>f
C. Fn<F<Fn+f
D. All

2. If a body of mass m is moving on rough horizontal surface of coefficient mu then the net electromagnetic force exerted by the surface on the body is -

Relevant Equations

Newton second law.

I don’t know what is contact force. Are friction and normal forces called contact forces? And we have to take the resultant of the two to get the net contact force?

 

[jbriggs444]

I don’t know what is contact force. Are friction and normal forces called contact forces? And we have to take the resultant of the two to get the net contact force?

I was going to provide a definition, but the one in Wikipedia is a bit better.

 

[rudransh verma]

I was going to provide a definition, but the one in Wikipedia is a bit better.

So to find the contact force we need to take the resultant of all the forces since that is what it is, the resultant of all the contact forces acting on body. Gravity is not a contact force but normal forces, and friction all come under contact forces.

What if there is applied force too acting on the body?

 

[jbriggs444]

So to find the contact force we need to take the resultant of all the forces since that is what it is, the resultant of all the contact forces acting on body. Gravity is not a contact force but normal forces, and friction all come under contact forces.

What if there is applied force too acting on the body?

What does Newton's second law say?

 

[rudransh verma]

What does Newton's second law say?

F=ma. I think if there is Friction there is applied force too. So sum of all the forces will give contact force or will have to leave applied force?

 

[jbriggs444]

F=ma. I think if there is Friction there is applied force too. So sum of all the forces will give contact force or will have to leave applied force?

Newton's second law states that $\sum F = ma$. Not one force. The sum of all forces.

There can be friction without an applied force. A block that slides to a stop, for instance.

 

[rudransh verma]

Newton's second law states that $\sum F = ma$. Not one force. The sum of all forces.

There can be friction without an applied force. A block that slides to a stop, for instance.

So what do you want to say ?

 

[jbriggs444]

So what do you want to say ?

I thought that I was clear. You said:

I think if there is Friction there is applied force too.

That was incorrect. There can be friction without some other applied force. An example is a block sliding to a stop.

 

[haruspex]

What if there is applied force too acting on the body?

The Wikipedia article gives "Pushing a car up a hill" as an example of a contact force.

 

[rudransh verma]

Newton's second law states that ∑F=ma. Not one force. The sum of all forces.

So a ramp applies a force known as contact force which can be decomposed into normal and friction. It is the force which is present between matter.

The Wikipedia article gives "Pushing a car up a hill" as an example of a contact force.

I don’t think that’s what it says. It says “where contact forces are at work”

 

[haruspex]

I don’t think that’s what it says. It says “where contact forces are at work”

It says: "Pushing a car up a hill ... are some of the everyday examples where contact forces are at work. In the first case the force is continuously applied by the person on the car".
So, yes, it does say pushing a car up a hill is an everyday example of a contact force.

A contact force acting on a body is any force that is applied to the surface by direct physical contact. The only forces I can think of that are not contact forces are gravity, electric and magnetic forces, and virtual (or "fictitious") forces such as centrifugal, inertial, Coriolis...

 

[jbriggs444]

So a ramp applies a force known as contact force which can be decomposed into normal and friction. It is the force which is present between matter.

Yes.

 

[rudransh verma]

Yes.

So, f> Fn.
And since F=under root(Fn^2+f^2)
So, F>Fn
And given ans is Fn+f>F. How?
Also F>f, how?

 

[jbriggs444]

So, f> Fn.
And since F=under root(Fn^2+f^2)
So, F>Fn
And given ans is Fn+f>F. How?
Also F>f, how?

Using variable names without saying what those variable names represent. You do this persistently. Please stop. Use your words. Tell us what the variable names represent.

You claim that "f > Fn".

Sure. If by "$f$" you mean the contact force and by "$F_n$" you mean the normal component of the contact force and if we are working on a ramp with a rough surface so that friction $F_f$ is present and is non-zero then the [magnitude of the] contact force $f$ will be greater than its normal component $F_n$.

But the next words out of your mouth are that $F=\sqrt{{F_n}^2+f^2}$.

Sure. If by "$F$" you mean the contact force, by "$f$" you mean friction and by "$F_n$" you mean the normal component of the contact force. But this is a different assignment of variable names than in the previous breath.

It is called the fallacy of equivocation.

 

[rudransh verma]

Sure. If by "f" you mean the contact force and by "Fn" you mean the normal component of the contact force and if we are working on a ramp with a rough surface so that friction Ff is present and is non-zero then the [magnitude of the] contact force f will be greater than its normal component Fn.

In question it’s already given f is friction , F is contact and Fn is normal. There is no mention in OP of ramp. I think because f=muFn. So f>Fn. But then I thought Mu is between 0 to 1. So I don’t think f>Fn . (This is not asked by the way)

But the next words out of your mouth are that F=Fn2+f2.

I am unable to mathematically prove how F>f or F>Fn.( I guessed!)

 

[haruspex]

f is friction , F is contact and Fn is normal

Just checking: you do understand that the question is saying that F consists of the two components f and Fn, yes?

f=muFn

You don't know that. If this is static friction then f is anything up to $-\mu_sF_n$.

I thought Mu is between 0 to 1.

There is no limit to the value of a coefficient of friction, though most are less than one.

 

[rudransh verma]

Just checking: you do understand that the question is saying that F consists of the two components f and Fn, yes?

Ok! So how do you prove F>f and F>Fn mathematically?

 

[jbriggs444]

Ok! So how do you prove F>f and F>Fn mathematically?

It is seldom educational to prove trivialities.

So you want to prove that the magnitude of the total force $|\vec{F}|$ is greater than both its the magnitude of its frictional component ($|f|$) and the magnitude of its normal component ($|F_n|$), right?

We can take it as a given that $|\vec{F}| = \sqrt{f^2 + {F_n}^2}$

The square root function is strictly monotone increasing. For any non-negative $x$ and $y$ such that $x > y$, it follows that $\sqrt{x} > \sqrt{y}$.

If $f$ (friction) is non-zero, we know that $f^2$ is strictly positive.
If $F_n$ (normal force) is non-zero, we know that ${F_n}^2$ is strictly positive.

It follows both that $f^2 + {F_n}^2 > f^2$ and that $f^2 + {F_n}^2 > {F_n}^2$

From which it follows both that $|F| = \sqrt{f^2 + {F_n}^2} > \sqrt{f^2} = |f|$ and that $|F| = \sqrt{f^2 + {F_n}^2} > \sqrt{{F_n}^2} = |F_n|$

However, there is a loophole in the proof. If $f$ is zero (frictionless) then the magnitude of the total force will not be greater than the magnitude of the normal component. Instead, the two will be equal.

Similarly, if $F_n$ is zero (no normal force) then the magnitude of the total force will not be greater than the magnitude of the frictional component. Instead, the two will be equal.

 

[rudransh verma]

It is seldom educational to prove trivialities.

What does that mean?

If f (friction) is non-zero, we know that f2 is strictly positive.
If Fn (normal force) is non-zero, we know that Fn2 is strictly positive.

It follows both that f2+Fn2>f2 and that f2+Fn2>Fn2

From which it follows both that |F|=f2+Fn2>f2=|f| and that |F|=f2+Fn2>Fn2=|Fn|

Ho

So F>Fn and F>f. Also Fn<Fn+f. So clearly All of these.
But can we also prove F<Fn+f ?
I have started proving: $$-F<-f$$
$$-F<-Fn$$
Adding both $$-F-F<-Fn-f$$
$$-2F<-Fn-f$$
$$2F>Fn+f$$
Can we proceed further?

 

[jbriggs444]

What does that mean?

If you ask for a proof that 2+2 = 4, the result will not be educational. Depending on what you accept as axiomatic, it can be either trivial and uninformative or quite lengthy and ultimately boring.

So F>Fn and F>f. Also Fn<Fn+f. So clearly All of these.
But can we also prove F<Fn+f ?

Huh?

$\vec{F}$ (the total force) is, by definition equal to the sum of its components, $\vec{F_n}$ and $\vec{f}$.

Possibly you want a proof of the triangle inequality? Given the lengths of two sides of a triangle, the length of the third side is no greater than the sum of the other two.

 

[rudransh verma]

you ask for a proof that 2+2 = 4, the result will not be educational. Depending on what you accept as axiomatic, it can be either trivial and uninformative or quite lengthy and ultimately boring.

Ok!
But why is F not equal to muFn always and what are the values of mu ? Is it not always from 0 to 1?
Because I recently learned otherwise.

 

[Steve4Physics]

But why is F not equal to muFn always

Because there is no reason it should be. For example: what if there is no friction but there is a normal force?

and what are the values of mu ? Is it not always from 0 to 1?

There is no upper limit on $\mu_s$ (coefficient of limiting static friction) or $\mu_k$ (coefficient of kinetic friction). @haruspex has already told you this (Post #16).

By the way, $\mu_sF_n$ gives the frictional force between 2 surfaces when slipping is about to start.
If there is no slipping, the frictional force could be less than this will be less than or equal to this.

Edited.

 

[haruspex]

why is F not equal to muFn always

For kinetic friction, $|F_{k}|=\mu_kN$.
For static friction, $|F_{s}|<=\mu_sN$.
If a body is resting on a frictional surface and a horizontal force $F_h$ is gradually applied then while $|F_{h}|<\mu_sN$ there is no movement. Therefore $F_s=-F_h$.

Please read https://www.physicsforums.com/insights/frequently-made-errors-mechanics-friction/.

Because I recently learned otherwise.

Which, that $F_s$ always equals $\mu_sN$ or that $\mu<=1$, or both?
Where did you learn that from?
Long list of coefficients at https://www.engineeringtoolbox.com/friction-coefficients-d_778.html shows some exceed 1.

 

[rudransh verma]

Because there is no reason it should be. For example: what if there is no friction but there is a normal force?

For second question in OP, by electromagnetic force do they mean contact force?
Because friction and normal forces all arise through electromagnetism between atoms and molecules.

 

[haruspex]

For second question in OP, by electromagnetic force do they mean contact force?
Because friction and normal forces all arise through electromagnetism between atoms and molecules.

They arise from electrostatic interactions between electrons in atoms, but those are still surface effects, so count as contact forces. More generally, electromagnetic interactions may be more long range, reaching to the interiors, and then do not count as contact forces.

 

[rudransh verma]

They arise from electrostatic interactions between electrons in atoms, but those are still surface effects, so count as contact forces. More generally, electromagnetic interactions may be more long range, reaching to the interiors, and then do not count as contact forces.

Edit: $$F_c=\sqrt{{(mg)}^2+{(mumg)}^2}$$
$$F_c=mg\sqrt{1+{(mu)}^2}$$

 

[jbriggs444]

$$F_c=\sqrt{{mg}^2+{mumg}^2}$$
$$F_c=mg\sqrt{1+{mu}^2}$$

What does $mumg^2$ mean? Are you trying to write $(\mu mg)^2$, aka "$(\text{mu}\ mg)^2$"

What you have comes out looking like it means $um^2g^2$.

You also have a problem with $mg^2$ where $(mg)^2$ is intended.

Curly braces in the $\LaTeX$ source are formatting hints only. They do not render in the finished product.

 

[Steve4Physics]

$$F_c=mg\sqrt{1+{mu}^2}$$

So you are saying $F_c=mg\sqrt{1+{\mu}^2}$.

Suppose you weigh (mg =) 700N. You are standing (in equilibrium) on level ground. The coefficient of (static) friction between you and the ground is $\mu_s=0.5$.

What is the contact force $F_c$ acting on you?

Hint: it is not $700\sqrt{1+0.5^2}$.

 

[rudransh verma]

What is the contact force Fc acting on you?

Well It should be 700N since friction is zero(Standing), not $mumg$

 

[kuruman]

Well It should be 700N since friction is zero(Standing), not $mumg$

OK, that is correct. Now suppose you weigh 700 N and you stand at rest on a rough plane inclined at 12° relative to the horizontal. What is the contact force on you in this case? Give magnitude and direction.

 

[Steve4Physics]

Well It should be 700N since friction is zero(Standing), not $mumg$

Yes, 700N. But putting the values into your formula gives $F_c = 700\sqrt{1+0.5^2} = 783N$ approx, which is clearly wrong.

Call the normal reaction $F_n$ (you are using $F_n=mg$).

It is important to realize that $\mu_sF_n$ does not give the frictional force except in special circumstances.

Do you know under what special circumstances the frictional force is $\mu_sF_n$?

(By the way, I think @kuruman is raising a different, but important, issue in Post #30.)

And note that to get the Greek letter $\mu$ in LaTeX, use \mu.

 

[rudransh verma]

OK, that is correct. Now suppose you weigh 700 N and you stand at rest on a rough plane inclined at 12° relative to the horizontal. What is the contact force on you in this case? Give magnitude and direction.

3796.15 N
90.5 degrees

 

[jbriggs444]

3796.15 N
90.5 degrees

Simpler than that. Way simpler.

You have a 700 N man exerting a 3796.15 N force. Nice trick if you can pull it off.

 

[kuruman]

3796.15 N
90.5 degrees

Huh? Think again. This is a teaching moment. There are two forces acting on you, your weight exerted by the Earth and the contact force exerted by the incline. They add up to zero because you are rest and remain at rest. So $\dots$

 

[rudransh verma]

Huh? Think again. This is a teaching moment. There are two forces acting on you, your weight exerted by the Earth and the contact force exerted by the incline. They add up to zero because you are rest and remain at rest. So $\dots$

The contact force namely the resultant of friction and normal should be same and opposite to our weight. So 700N opposite of gravity.
We can also say the vector sum of gravity, normal and friction should be zero for the body to remain at rest.

 

[kuruman]

The contact force namely the resultant of friction and normal should be same and opposite to our weight. So 700N opposite of gravity.
We can also say the vector sum of gravity, normal and friction should be zero for the body to remain at rest.

Very good. You got it. There is only one contact force exerted by the incline and it has components like any other force. Its component perpendicular to the incline is also called the normal force and its component parallel to the incline is also called friction.

 

[rudransh verma]

Very good. You got it. There is only one contact force exerted by the incline and it has components like any other force. Its component perpendicular to the incline is also called the normal force and its component parallel to the incline is also called friction.

But why I got the wrong result. I broke the three forces in components and added the all vertical components. Similarly added all the horizontal ones. Got two eqns. Solved. But got Fn and f such that the resultant is some way more than 700N.

 

[kuruman]

But why I got the wrong result. I broke the three forces in components and added the all vertical components. Similarly added all the horizontal ones. Got two eqns. Solved. But got Fn and f such that the resultant is some way more than 700N.

You have posted here enough times to know that asking why you got the wrong result is insufficient. You have to show us what you did. Even though we often do a decent job as homework helpers, we suck consistently as mind readers.

 

[rudransh verma]

You have posted here enough times to know that asking why you got the wrong result is insufficient. You have to show us what you did. Even though we often do a decent job as homework helpers, we suck consistently as mind readers.

$F_N\cos78-f\cos12=0$
$F_N\sin78+f\sin12-700=0$
After solving,
$f=744 N$
$F_N=3638.5 N$

 

[haruspex]

$F_N\cos78-f\cos12=0$
$F_N\sin78+f\sin12-700=0$
After solving,
$f=744 N$
$F_N=3638.5 N$

Let's check if that satisfies your equations:
$3638.5\sin(78°)+744\sin(12°)=3556+155=3711$.
Not 700.
You could, and should, have done that check for yourself.
Please post your working if you cannot find your mistake.

 

[rudransh verma]

Let's check if that satisfies your equations:
$3638.5\sin(78°)+744\sin(12°)=3556+155=3711$.
Not 700.
You could, and should, have done that check for yourself.
Please post your working if you cannot find your mistake.

I got it! I was not multiplying by $\cos78$ to right hand side of second eqn.
I got $F_N=762.5N$
$f=155.5N$
$F_C=778.19N$ (maybe some small differences in decimal)

 

[haruspex]

I got it! I was not multiplying by $\cos78$ to right hand side of second eqn.
I got $F_N=762.5N$
$f=155.5N$

Does that satisfy your equations?

 

[rudransh verma]

Does that satisfy your equations?

No! But I know it’s because I have rounded in the middle of solution. Small difference of 6 N in first eqn.

 

[haruspex]

No! But I know it’s because I have rounded in the middle of solution. Small difference of 6 N in first eqn.

Your error is rather large, more than 10%. Please post your working.

 

[jbriggs444]

I got it! I was not multiplying by $\cos78$ to right hand side of second eqn.
I got $F_N=762.5N$
$f=155.5N$
$F_C=778.19N$ (maybe some small differences in decimal)

You are trying to come up with two components which combine (via the Pythagorean theorem) to produce a sum of 700 N.

It is immediately obvious that neither component can be greater than their vector sum.

 

[rudransh verma]

Please post your working.

$F_N\cos78-f\cos12=0$
$F_N\sin78+f\sin12=700$
So multiplying by sin78 in first and by cos78 in second eqn
$F_N\cos78\sin78-f\cos12\sin78=0$
$F_N\sin78\cos78+f\sin12cos78=700cos78$
On putting values of all trigonometric functions
$F_N(0.2)(0.97)-f(0.97)(0.97)=0$
$F_N(0.97)(0.2)+f(0.2)(0.2)=700(0.2)$

$f=142.72N$ and $F_N=692.19N$
Now its less than the resultant force.
But still there is a difference in the values.

 

[jbriggs444]

$F_N\cos 78-f\cos 12=0$
$F_N\sin 78+f\sin 12=700$
So multiplying by $\sin 78$ in first and by $\cos 78$ in second eqn
$F_N\cos 78 \sin 78-f \cos12 \sin 78=0$
$F_N\sin 78 \cos 78+f \sin 12 \cos 78=700 \cos 78$
On putting values of all trigonometric functions
$F_N(0.2)(0.97)-f(0.97)(0.97)=0$
$F_N(0.97)(0.2)+f(0.2)(0.2)=700(0.2)$

[after minor repairs on your $\LaTeX$]

Were you actually going to try to solve those equations?

The obvious next step would be to subtract the last two equations, thus arriving at an equation for $f$ alone. But if you were going to do that, it would be premature to substitute in values for the trig functions.

Although there are easier ways to break a vector into components, one can indeed do the job with simultaneous equations. Crucially, one must solve the simultaneous equations.

 

[kuruman]

Crucially, one must solve the simultaneous equations.

Indeed.
To @rudransh verma :
It would be best to not substitute numbers before solving because they often hide what is actually going and ho to take the next algebraic step. We have,
$F_N\sin\!\theta-f\cos\!\theta = 0~~~~~~~~~~(1)$
$F_N\cos\!\theta+f\sin\!\theta = W~~~~~~~~(2)$

I will follow your method and multiply the first equation by $\cos\!\theta$ and the second by $\sin\!\theta$
$F_N\sin\!\theta\cos\!\theta-f\cos^2\!\theta = 0~~~~~~~~~~~~~~~~~(3)$
$F_N\cos\!\theta\sin\!\theta+f\sin^2\!\theta = W\sin\!\theta~~~~~~~~(4)$

As @jbriggs444 suggested, subtract equation (3) from (4) to get
$f\sin^2\!\theta+f\cos^2\!\theta=W\sin\!\theta~\implies f=W\sin\!\theta$
Finally, put this value for $f$ in either (1) or (2) to get $F_N=W\cos\!\theta$,

You have been unable to get a reasonable answer. Why? My answer to that is that you have not yet acquired the single good habit that would limit algebraic and arithmetic mistakes and round off errors while simultaneously making it easy to troubleshoot your workings in case there is an actual mistake. That good habit is not substituting numbers until the very end.

What happened here? Not only you substituted numbers, but also you made things worse by mixing values of angles, using both 12° and 78°. I am sure you know that cos12° = sin78° but I am not sure that you would recognize on sight that (cos12°/sin78°) and (cos²12° + sin²78°) are both equal to 1.

I hope I have convinced you that your insistence on substituting numbers early on is actually a barrier to your ability to work out physics problems and a waste of your time. Using symbols instead of numbers makes it easier to be correct the first time around. I know that you been told that before, but here you have proof, three unsuccessful attempts at a solution, that your insistence on doing it your way does not work as well as you would want.