Finding the angle when given the coefficient of frictionby NYILoveYou Tags: angle, coefficient, friction 

#1
Feb1111, 04:01 PM

P: 7

1. The problem statement, all variables and given/known data
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. 2. Relevant equations The coefficient of friction = Friction/ Normal Force Pythagoras Theorem 3. 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. 



#2
Feb1111, 04:25 PM

Mentor
P: 40,907

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
Feb1211, 05:10 AM

P: 460

Hint: you are missing a major point. there is a force holding the mop. otherwise the mop will fall.




#4
Feb1211, 05:11 AM

P: 460

Finding the angle when given the coefficient of friction
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..




#6
Feb1211, 12:41 PM

P: 7

I'm not too sure on the vertical component as you don't actually have the opposite side? 



#7
Feb1211, 04:29 PM

Mentor
P: 40,907

Trigonometric Method of Vector Resolution Resolving a Vector Into Components 



#8
Feb1211, 05:07 PM

P: 7

so the horizontal component is Fcosθ the vertical component would be Fsinθ i still don't see how that will help, as you don't actually have F or θ 



#9
Feb1211, 05:11 PM

Mentor
P: 40,907





#10
Feb1411, 03:15 PM

P: 7

Fr=coefficient of friction x the normal force? 



#11
Feb1411, 05:32 PM

Mentor
P: 40,907





#12
Feb1511, 11:19 AM

P: 7

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
Feb1511, 11:31 AM

Mentor
P: 40,907





#14
Feb1911, 01:41 PM

P: 7

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
Feb1911, 02:15 PM

Mentor
P: 40,907

So far, so good. Can you replace that ratio of trig functionsor its inversewith a single trig function? Then use your calculator.




#16
Feb1911, 03:55 PM

P: 7

THANKS!! I finally did it! Thank you so much! :D



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