# Finding the angle when given the coefficient of friction

by NYILoveYou
Tags: angle, coefficient, friction
 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.
Mentor
P: 41,579
 Quote by NYILoveYou 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.
 P: 460 Hint: you are missing a major point. there is a force holding the mop. otherwise the mop will fall.
 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..
Mentor
P: 41,579
 Quote by ashishsinghal 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.
P: 7
 Quote by Doc Al 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?
Mentor
P: 41,579
 Quote by NYILoveYou Well the horizontal force would be cos θ = 0.3/F
The horizontal component of the force F would be Fcosθ.

 I'm not too sure on the vertical component as you don't actually have the opposite side?
You may need to review how to find the components of a vector. See:
Trigonometric Method of Vector Resolution
Resolving a Vector Into Components
P: 7
 Quote by Doc Al The horizontal component of the force F would be Fcosθ. You may need to review how to find the components of a vector. See: Trigonometric Method of Vector Resolution Resolving a Vector Into Components
Thanks, I remember now,
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
Mentor
P: 41,579
 Quote by NYILoveYou 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?
P: 7
 Quote by Doc Al 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?
Mentor
P: 41,579
 Quote by NYILoveYou 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!
P: 7
 Quote by Doc Al 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?
Mentor
P: 41,579
 Quote by NYILoveYou would it be... Fcosθ = 0.3 x Fsinθ?
Looks good. Now just solve for θ. (Rearrange a bit and reach for your calculator.)
 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?
 Mentor P: 41,579 So far, so good. Can you replace that ratio of trig functions--or its inverse--with a single trig function? Then use your calculator.
 P: 7 THANKS!! I finally did it! Thank you so much! :D

 Related Discussions Introductory Physics Homework 1 Introductory Physics Homework 11 Introductory Physics Homework 3 Introductory Physics Homework 5 Introductory Physics Homework 3