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

In summary: I think I got it now. I was trying to get rid of the 0.295F and still solve for F. I forgot to use the 0.705F that actually cancels out the 0.295F. So now if I divide both sides by 0.755 I get F=6.88. Thank you so much!...I think I got it now. I was trying to get rid of the 0.295F and still solve for F. I forgot to use the 0.705F that actually cancels out the 0.295F. So now if I divide both sides by 0.755 I get F=6.88. Thank you so much!
  • #1
Schaus
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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
 
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  • #2
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.
 
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  • #3
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
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##?
 
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  • #5
I thought that if I put the (10.78)+(Sin41) in my calculator that I have my Fn now?
 
  • #6
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##
 
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  • #7
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
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.
 
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  • #9
(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.
 
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  • #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
 
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  • #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##.
 
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  • #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.
 
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  • #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?
 
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  • #18
Oh wow that was a bit stupid on my part! It makes sense now! Thanks for all your help, really appreciate it!
 

1. How do you calculate downward force for a curling broom at constant speed?

In order to calculate the downward force for a curling broom at constant speed, you will need to use the formula F=ma, where F represents force, m represents mass, and a represents acceleration. First, determine the mass of the broom and the acceleration it is experiencing while being used at a constant speed. Then, multiply the mass by the acceleration to find the downward force.

2. What is the significance of the F=ma equation in calculating downward force for a curling broom?

The F=ma equation is derived from Newton's second law of motion, which states that the force applied to an object is directly proportional to its mass and acceleration. In the case of a curling broom, this equation helps us understand the relationship between the force being applied to the broom, its mass, and how quickly it is accelerating. By using this equation, we can accurately calculate the downward force on the broom at a constant speed.

3. How does the downward force for a curling broom at a constant speed affect the movement of the broom?

The downward force on a curling broom at a constant speed is what helps to keep the broom in contact with the ice. This force is essential in allowing the sweeping motion to effectively manipulate the path of the curling stone. Without enough downward force, the broom may not be able to maintain proper contact with the ice, resulting in less effective sweeping.

4. Can the downward force for a curling broom at constant speed be altered?

Yes, the downward force for a curling broom at constant speed can be altered by changing either the mass or the acceleration of the broom. Increasing the mass of the broom or decreasing the acceleration will result in a higher downward force, while decreasing the mass or increasing the acceleration will result in a lower downward force.

5. Are there any other factors that may affect the downward force for a curling broom at constant speed?

Yes, there are other factors that may affect the downward force for a curling broom at constant speed. These include the quality and texture of the ice, the angle of the broom, and the strength and technique of the sweeper. All of these elements can influence the amount of downward force being applied to the broom and therefore affect the effectiveness of the sweeping.

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