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Finding the angle when given the coefficient of friction

  • Thread starter NYILoveYou
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  • #1

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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Answers and Replies

  • #2
Doc Al
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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.
 
  • #3
Hint: you are missing a major point. there is a force holding the mop. otherwise the mop will fall.
 
  • #4
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..
 
  • #5
Doc Al
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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.
 
  • #6
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?
 
  • #7
Doc Al
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  • #8
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  • #9
Doc Al
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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
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
Doc Al
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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
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
Doc Al
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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
Doc Al
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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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