# Newtons' Second Law with Kinetic Friction Problem

#### SpacemanRich

1. The problem statement, all variables and given/known data

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.

2. Relevant equations
F = ma, fk = μk Fn

3. 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.

Rich.

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#### Trevorman

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

#### CWatters

Homework Helper
Gold Member
Out of interest is the book answer about 22 degrees?

#### Trevorman

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.

#### SpacemanRich

Out of interest is the book answer about 22 degrees?
The actual answer given is 68o

#### Trevorman

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 $tan(\theta) = \frac{1}{\mu_k} \rightarrow \theta = tan^{-1}(\frac{1}{\mu_k})$

#### SpacemanRich

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.

#### SpacemanRich

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.

#### Trevorman

$F_y=F*sin(\theta)$
$F_x=F*cos(\theta)$
$F_n-F_y = 0$ 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 $F_x$.

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

#### SpacemanRich

$F_y=F*sin(\theta)$
$F_x=F*cos(\theta)$
$F_n-F_y = 0$ 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 $F_x$.

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???

#### AlephNumbers

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.

#### CWatters

Homework Helper
Gold Member
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..

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?

#### SpacemanRich

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.

#### SpacemanRich

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??

#### Trevorman

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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#### SpacemanRich

"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.

#### AlephNumbers

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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#### SpacemanRich

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.

#### SpacemanRich

"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. Thanks so much.

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