# Force Problem

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1. Dec 23, 2016

### Schaus

1. The problem statement, all variables and given/known data
Barney is practicing his sweeping techniques for an upcoming curling tournament. He exerts a force on a 1.1-kg broom as he walks across a tile floor at a constant speed. The coefficient of friction between the floor and the broom is 0.45 and the broom handle makes an angle of 41° with the horizontal. Determine the amount of force with which Barney pushes downward (along the handle of the broom) in order to achieve this constant speed motion. Begin with a free body diagram. (ans: 10.6N)

2. Relevant equations
F=ma

3. The attempt at a solution
I created a FBD and using the info from the question I came up with
(10.78N)+(Fsin41°)=Fnorm
Fn=11.44N
(0.45)(Fn)=Fcos41°
F=6.82N now if this is my horizontal force then it will be the same as my friction force since the broom isn't accelerating.
If I want the force on the broom then using SOH CAH TOA. I take A/cos41 = 9.03N. I'm at a loss of how to find 10.6N

2. Dec 23, 2016

### PeroK

I think you are correct until the end. You have:

$0.45F_n = F \cos(\theta)$

What did you do after that?

Correction: how did you get $F_n = 6.82N$? That can't be right.

3. Dec 23, 2016

### Schaus

Well if (10.78)+(sin41)=11.44N then
(0.45)(11.44)=Fcos41
5.148=Fcos41 divide both sides by cos41
F=6.82N

4. Dec 23, 2016

### PeroK

How do you know the normal force is $11.44N$?

5. Dec 23, 2016

### Schaus

I thought that if I put the (10.78)+(Sin41) in my calculator that I have my Fn now?

6. Dec 23, 2016

### PeroK

Where did the $F$ disappear to? It should be $F \sin\theta$

7. Dec 23, 2016

### Schaus

Oh woops! Ok so now if I sub (10.78N)+(Fsin41°)=Fnorm
(0.45)(Fn)=Fcos41°
(0.45)(10.78N)+(Fsin41°)=Fcos41°
But now if I move the Fsin41 over to Fcos41 won't that cancel out my F?

8. Dec 23, 2016

### PeroK

The coeff of friction applies to all of $F_n$. In any case, you just need a bit algebra now.

9. Dec 23, 2016

### Schaus

(0.45)(10.78N)+(Fsin41°)=Fcos41°
4.851N=Fcos41°-Fsin41° won't this cancel out my F's though? Meaning I have no variable anymore?

10. Dec 23, 2016

### PeroK

That's still wrong. The coeff of friction applies to the $F\sin\theta$. In any case, I don't see why $F$ would cancel out of the equation.

11. Dec 23, 2016

### Schaus

Sorry, I'm a bit lost. I'll try to show you what I think you mean.
(0.45)(10.78N)+(Fsin41°)=Fcos41°
4.851+(Fsin41°)=Fcos41°
4.851+(F(0.656))=F(0.755)
5.507N=F(0.755)
7.29N = F
I'm not sure what I'm doing wrong.

12. Dec 23, 2016

### TomHart

The first line is right, but the second line is wrong. Just an algebra error.
It should be: (0.45)(10.78) + (0.45)(Fsin41) = Fcos41

13. Dec 23, 2016

### PeroK

You were doing all right in your first post. You got:

$F_n = mg + F \sin\theta$

$F_f = 0.45F_n$

Where $F_f$ is the frictional force. And:

$F_f = F \cos \theta \$ (for equilibrium)

You had all that. But, then you went wrong. You should have got:

$F \cos \theta = 0.45F_n = 0.45 ( mg + F \sin\theta)$

You just applied the $0.45$ to the first term in the normal force. You must apply it to all of the normal force.

To finish this off, you can in fact just plug in all the numbers and you'll get a numeric equation for $F$.

14. Dec 23, 2016

### Schaus

Ok I understand my error with the co-efficient of friction now. But now I'm getting 6.82N again...sorry, I'll show you what I'm doing
(0.45)(10.78) + (0.45)(Fsin41) = Fcos41
4.851 + 0.295 = Fcos41
5.15 = Fcos41 if I divide both sides by cos 41 I get 6.82 again
6.82N = F

15. Dec 23, 2016

### PeroK

You've just dropped one of the $F's$ out of the equation. I've highlighted it in bold above.

16. Dec 23, 2016

### Schaus

I'm sorry this just doesn't seem to be clicking.
(0.45)(10.78) + (0.45)(Fsin41) = Fcos41
4.851 + 0.295F = Fcos41
-0.295F = 0.705Fcos41
4.851 = 0.705F cos41 divide both sides by .705 then cos 41 and I get 9.12N
6.88 = Fcos41
9.12 = F
I hope I'm at least going in the right direction and I really appreciate both your help!

17. Dec 23, 2016

### PeroK

I'm not sure why you calculated $sin(41)$ but not $cos(41) = 0.755$. So, you should have:

$4.851 + 0.295F = 0.755F$

Can you finish it off from there?

18. Dec 23, 2016

### Schaus

Oh wow that was a bit stupid on my part! It makes sense now! Thanks for all your help, really appreciate it!