Newtons' Second Law with Kinetic Friction Problem

In summary: Yes, I'm familiar with the concept of limits. Unfortunately, I don't think that would be of any help here.
  • #1
SpacemanRich
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2

Homework Statement



While moving in, a new homeowner is pushing a box across the floor at a constant velocity. The coefficient of kinetic friction between the box and the floor is 0.41. The pushing force is directed downward and an angle θ below the horizontal. When θ is greater than a certain value, it is not possible to move the box, no matter how large the pushing force is. Find the that value of θ.

Homework Equations


F = ma, fk = μk Fn

The Attempt at a Solution


I have tried working this problem multiple times, I must be missing something in the initial set up. I keep ending up with variables that don't cancel out, or I have given values for. (The only value given is for the coefficient for
kinetic friction). The solution given is just an angle in degrees.

I could really use a walk through of the solution to this problem. I'm missing some concept in my initial setup, and I just can't seem to see it.

Thanks in advance.
Rich.
 
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  • #2
What if you try to make a free body diagram, from that I get

## \mu_k ( F_{1} sin(\theta) + F_n) < F_1 cos(\theta) ##

Where ##F_1## is the pushing force, ##F_n## the normal force of the box
 
  • #3
Out of interest is the book answer about 22 degrees?
 
  • #4
CWatters said:
Out of interest is the book answer about 22 degrees?

Well with that little info that you have, the ## F_k = \mu_k F_n ##, the friction force divided with the normal force ## \frac{F_k}{F_n} = \mu_k ##, means that the angle will become ##\theta = tan^{-1}(\mu_k)##. I can give you a more detailed explanation in a while.
 
  • #5
CWatters said:
Out of interest is the book answer about 22 degrees?
The actual answer given is 68o
 
  • #6
SpacemanRich said:
The actual answer given is 68o

Sorry a miscalculation

## tan(\theta) = \frac{F_n}{F_k} = \frac{F*sin(\theta)}{F*cos(\theta)} = \frac{1}{\mu_k}##
From this equation you can see that ##tan(\theta) = \frac{1}{\mu_k} \rightarrow \theta = tan^{-1}(\frac{1}{\mu_k})##
 
  • #7
Trevorman said:
Well with that little info that you have, the ## F_k = \mu_k F_n ##, the friction force divided with the normal force ## \frac{F_k}{F_n} = \mu_k ##, means that the angle will become ##\theta = tan^{-1}(\mu_k)##. I can give you a more detailed explanation in a while.

This seems to be on the right track. The answer (by guessing, and trial and error) looks to be ##\theta = tan^{-1}(1/\mu_k)##.

The problem is I don't know why. Can you post your Free Body Diagram? I think that is where I made my mistake.

Thanks.
 
  • #8
Trevorman said:
Sorry a miscalculation

## tan(\theta) = \frac{F_n}{F_k} = \frac{F*sin(\theta)}{F*cos(\theta)} = \frac{1}{\mu_k}##
From this equation you can see that ##tan(\theta) = \frac{1}{\mu_k} \rightarrow \theta = tan^{-1}(\frac{1}{\mu_k})##

That's the same answer I guessed at. But I still don't understand Why? It would help to see the FBD.
 
  • #9
rrhwg6.jpg


##F_y=F*sin(\theta)##
##F_x=F*cos(\theta)##
##F_n-F_y = 0## Forces in y-direction
Due to the constant velocity the forces in the x-direction ##F_x-F_k = 0##
##\theta## is the angle between F and ##F_x##.

j8gl1f.jpg

The result force diagram will look like the picture above, from there it is just geometry.
 
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  • #10
Trevorman said:
rrhwg6.jpg


##F_y=F*sin(\theta)##
##F_x=F*cos(\theta)##
##F_n-F_y = 0## Forces in y-direction
Due to the constant velocity the forces in the x-direction ##F_x-F_k = 0##
##\theta## is the angle between F and ##F_x##.

j8gl1f.jpg

The result force diagram will look like the picture above, from there it is just geometry.

That looks "Almost" the same as mine. The only difference is that I included a vector for the Weight of the box. That also seems to be where my problem came from.
Can you tell me why you didn't include the weight of the box?

I'm pretty certain that is my mistake. I just don't understand why it should be excluded?
 
  • #11
Are you familiar with the concept of limits? I would also like to add that I think that you should include the weight of the box in the FBD, even if the weight of the box ultimately proves negligible.
 
  • #12
SpacemanRich said:
I'm pretty certain that is my mistake. I just don't understand why it should be excluded?

Not "should" but "can"...

There is a hint in..

SpacemanRich said:
no matter how large the pushing force is

One of the equations contains...

μk(mg + FSinθ)

What can you do to simplify that if F is extremely large compared to mg?
 
  • #13
AlephNumbers said:
Are you familiar with the concept of limits? I would also like to add that I think that you should include the weight of the box in the FBD, even if the weight of the box ultimately proves negligible.

Yes, I'm familiar with the concept of a Limit. Unfortunately, I don't think that would be of any help here.

This problem is from an Algebra based Intro to Physics text. Cutnell and Johnson, 9th edition to be exact.

So, I think we are looking for an Algebra based solution.
 
  • #14
CWatters said:
Not "should" but "can"...

There is a hint in..
One of the equations contains...

μk(mg + FSinθ)

What can you do to simplify that if F is extremely large compared to mg?

If F is extremely large it would make mg too small to matter. But I don't think that a large F is a requirement to the solution, It seems more of a way to emphasize that the angle is the critical issue. That's the way it reads to me, anyway.
Thoughts??
 
  • #15
SpacemanRich said:
If F is extremely large it would make mg too small to matter. But I don't think that a large F is a requirement to the solution, It seems more of a way to emphasize that the angle is the critical issue. That's the way it reads to me, anyway.
Thoughts??
"no matter how large the pushing force is" means that F is very large, this means that F >> mg, as a result ##mg+F*sin(\theta) \approx F*sin(\theta)##
 
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  • #16
Trevorman said:
"no matter how large the pushing force is" means that F is very large, this means that F >> mg, as a result ##mg+F*sin(\theta) \approx F##

Ok, I'm going to start fresh with this new way of reading the problem. I'll be back after I give it a try. Thanks for your input.
 
  • #17
SpacemanRich said:
Yes, I'm familiar with the concept of a Limit. Unfortunately, I don't think that would be of any help here.

This problem is from an Algebra based Intro to Physics text. Cutnell and Johnson, 9th edition to be exact.

So, I think we are looking for an Algebra based solution.

In the context of an algebra-based book then I suppose it is reasonable not to analytically evaluate the limit. But you must make a mathematical agreement on closeness in order to solve this problem. There is no simple algebraic solution.

Or maybe there is. It is certainly beyond me, however.
 
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  • #18
AlephNumbers said:
In the context of an algebra-based book then I suppose it is reasonable not to analytically evaluate the limit. But you must make a mathematical agreement on closeness in order to solve this problem. There is no simple algebraic solution.

Or maybe there is. It is certainly beyond me, however.

Thanks. I agree. At least I know now I wasn't missing any obvious physics concept I should be learning. That is reassuring.
Thanks again.
 
  • #19
Trevorman said:
"no matter how large the pushing force is" means that F is very large, this means that F >> mg, as a result ##mg+F*sin(\theta) \approx F*sin(\theta)##

I reworked the problem with your understanding of the meaning of "extra" information, and approaching it that way everything works out. Thanks for the help.
Also, I have never encountered a problem where I had to apply information that way to solve the problem. It was very eye opening experience. When it was done, and I was rereading it, it looks almost beautiful in how it leads to the solution. Up until this problem, everything I had encountered was solved with straight formula use, or algebraic manipulation. I feel like I really learned a lot by getting through this problem. I couldn't have done it without your help. :smile::smile::smile: Thanks so much.
 
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What is Newton's Second Law?

Newton's Second Law, also known as the Law of Force and Acceleration, states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass.

What is kinetic friction?

Kinetic friction is the force that acts against the motion of an object when it is in contact with another surface. It is caused by the microscopic irregularities on the two surfaces rubbing against each other.

How is Newton's Second Law applied to a problem involving kinetic friction?

In a problem involving kinetic friction, Newton's Second Law is used to calculate the net force acting on the object. The force of kinetic friction is also taken into account, which is equal to the coefficient of kinetic friction multiplied by the normal force.

What is the coefficient of kinetic friction?

The coefficient of kinetic friction is a dimensionless constant that represents the amount of force required to keep an object in motion on a surface. It depends on the types of material and the surface properties in contact.

How do you solve a problem involving Newton's Second Law with kinetic friction?

To solve a problem involving Newton's Second Law with kinetic friction, you would first draw a free body diagram to identify all the forces acting on the object. Then, you would use Newton's Second Law to calculate the net force and set it equal to the product of mass and acceleration. Lastly, you would use the coefficient of kinetic friction to calculate the force of kinetic friction and add it to the net force to solve for the acceleration of the object.

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