Calculating Downward Force for Curling Broom at Constant Speed | F=ma Equation

  • Thread starter Thread starter Schaus
  • Start date Start date
  • Tags Tags
    Force Physics
Click For Summary

Homework Help Overview

The problem involves calculating the downward force exerted by a person on a curling broom while maintaining a constant speed across a surface. The scenario includes a broom with a specified mass, a coefficient of friction, and an angle of the broom handle with respect to the horizontal. Participants are tasked with determining the force required to achieve this motion using the principles of physics, particularly Newton's second law (F=ma).

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the setup of free body diagrams and the relationships between forces acting on the broom, including normal force and frictional force. There are attempts to derive equations based on the given parameters, with some participants questioning the accuracy of their calculations and the application of the coefficient of friction.

Discussion Status

There is an ongoing exploration of the relationships between the forces involved, with participants providing corrections and suggestions to each other. Some participants express confusion over their calculations, particularly regarding the normal force and the application of the coefficient of friction. Guidance has been offered to clarify the algebraic relationships, but no consensus has been reached on the final value of the force.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available and the methods they can use. There is a focus on ensuring that all forces are accounted for correctly, particularly in relation to the angle of the broom handle and the forces acting on it.

Schaus
Messages
118
Reaction score
5

Homework Statement


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)

Homework Equations


F=ma

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
 
Physics news on Phys.org
Schaus said:

Homework Statement


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)

Homework Equations


F=ma

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

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.
 
  • Like
Likes   Reactions: 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
 
Schaus said:
Well if (10.78)+(sin41)=11.44N then
(0.45)(11.44)=Fcos41
5.148=Fcos41 divide both sides by cos41
F=6.82N

How do you know the normal force is ##11.44N##?
 
  • Like
Likes   Reactions: Schaus
I thought that if I put the (10.78)+(Sin41) in my calculator that I have my Fn now?
 
Schaus said:
I thought that if I put the (10.78)+(Sin41) in my calculator that I have my Fn now?

Where did the ##F## disappear to? It should be ##F \sin\theta##
 
  • Like
Likes   Reactions: 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?
 
Schaus said:
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?

The coeff of friction applies to all of ##F_n##. In any case, you just need a bit algebra now.
 
  • Like
Likes   Reactions: 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
Schaus said:
(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?

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.
 
  • Like
Likes   Reactions: Schaus
  • #11
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
Schaus said:
(0.45)(Fn)=Fcos41°
(0.45)(10.78N)+(Fsin41°)=Fcos41°
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
 
  • Like
Likes   Reactions: Schaus
  • #13
Schaus said:
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.

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##.
 
  • Like
Likes   Reactions: Schaus
  • #14
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
Schaus said:
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 + 0.295F = Fcos41
5.15 = Fcos41 if I divide both sides by cos 41 I get 6.82 again
6.82N = F

You've just dropped one of the ##F's## out of the equation. I've highlighted it in bold above.
 
  • Like
Likes   Reactions: Schaus
  • #16
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
Schaus said:
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!

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?
 
  • Like
Likes   Reactions: Schaus
  • #18
Oh wow that was a bit stupid on my part! It makes sense now! Thanks for all your help, really appreciate it!
 

Similar threads

High School Understanding F=ma: Constant Speed & Acceleration
  • · Replies 9 ·
Replies
9
Views
2K
Minimum Force to Move a Crate with Constant Speed?
  • · Replies 8 ·
Replies
8
Views
13K