Finding the angle when given the coefficient of friction

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Homework Help Overview

The problem involves determining the angle between the handle of a mop and the horizontal, given a coefficient of friction of 0.3 while the mop is moving at a constant velocity. The context includes forces acting on the mop, such as gravity, normal force, and friction.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the forces acting on the mop, including gravity and friction, and question the assumption of negligible mass. There are attempts to express the relationship between the forces and the angle using trigonometric functions.

Discussion Status

Several participants have provided hints and corrections regarding the forces involved and the relationships between them. There is ongoing exploration of the equations relating the applied force, friction, and the angle, with some participants expressing uncertainty about the next steps.

Contextual Notes

Participants are navigating through assumptions about mass and forces, and there is an emphasis on understanding the balance of forces when the mop is in motion at a constant velocity.

NYILoveYou
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Homework Statement


A mop is being used to clean a floor. The coefficient of friction between the mop and floor is 0.3. Find the angle between the handle and the horizontal when the mop head is moving at constant velocity.


Homework Equations


The coefficient of friction = Friction/ Normal Force
Pythagoras Theorem


The Attempt at a Solution


I figured, the main force would be gravity at 9.8 as mass is negligible. I also tried to get the components of the force. If the downward force was 9.8m/s, the horizontal would be 0.3*9.8 - 2.94 (?). The normal force would also be 9.8, as there is not vertical acceleration or velocity.

I thought could you say perhaps, 2.94/9.8 and the inverse sine of this? It does give the right answer but it feels like the working is incorrect..

Please help, any pointers would be appreciated.
 
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NYILoveYou said:
I figured, the main force would be gravity at 9.8 as mass is negligible.
If the mass is negligible, then so is the force of gravity.

Do this: Assume that some force is being applied along the line of the mop handle. Call it F. Call the angle that the mop makes with the floor θ. Now figure out the horizontal and vertical forces on the mop.
 
Hint: you are missing a major point. there is a force holding the mop. otherwise the mop will fall.
 
no...the mass is not negligible. if that is so then gravity, normal and friction force would all be negligible. consider all forces and apply the concept of torque..
 
ashishsinghal said:
no...the mass is not negligible. if that is so then gravity, normal and friction force would all be negligible. consider all forces and apply the concept of torque..
That is incorrect.
 
Doc Al said:
If the mass is negligible, then so is the force of gravity.

Do this: Assume that some force is being applied along the line of the mop handle. Call it F. Call the angle that the mop makes with the floor θ. Now figure out the horizontal and vertical forces on the mop.

Well the horizontal force would be cos θ = 0.3/F

I'm not too sure on the vertical component as you don't actually have the opposite side?
 
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NYILoveYou said:
so the horizontal component is Fcosθ
the vertical component would be Fsinθ
Good.
i still don't see how that will help, as you don't actually have F or θ
Those aren't the only forces acting on the mop. What about the normal force and friction?
 
  • #10
Doc Al said:
Good.

Those aren't the only forces acting on the mop. What about the normal force and friction?

The normal force would be equal to friction, as the mop is moving at constant velocity,
Fr=coefficient of friction x the normal force?
 
  • #11
NYILoveYou said:
The normal force would be equal to friction, as the mop is moving at constant velocity,
Not exactly. Since, the velocity is constant the net force is zero. That means that the horizontal component of the applied force (F) must equal the friction force.
Fr=coefficient of friction x the normal force?
Right!
 
  • #12
Doc Al said:
Not exactly. Since, the velocity is constant the net force is zero. That means that the horizontal component of the applied force (F) must equal the friction force.

Right!

Thanks, so if

Fcosθ = Friction

And if friction = the coefficient of friction x the reaction force,

and the reaction force (R) = Fsinθ

would it be...

Fcosθ = 0.3 x Fsinθ?
 
  • #13
NYILoveYou said:
would it be...

Fcosθ = 0.3 x Fsinθ?
Looks good. Now just solve for θ. (Rearrange a bit and reach for your calculator.)
 
  • #14
Well I tried that, but then I got stuck...

The F's cancel out.

Move the sinθ to the other side, so you get

Cosθ/Sinθ = 0.3

then I got stuck, I was thinking perhaps you could use trigonometric proofs?
 
  • #15
So far, so good. Can you replace that ratio of trig functions--or its inverse--with a single trig function? Then use your calculator.
 
  • #16
THANKS! I finally did it! Thank you so much! :D
 

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