Help with friction Not sure where to start

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The discussion revolves around calculating the force required to push down on a mop handle to achieve constant velocity without acceleration. The handle makes an angle of 39.6° with the horizontal, and the mop has a mass of 3.07 kg, with a kinetic friction coefficient of 0.386. Participants clarify that the force of friction is dependent on the normal force, which increases when pushing down on the handle. There is confusion about how to apply the equations for friction and gravity, particularly regarding the orientation of forces involved. Understanding the relationship between the normal force and the downward force applied is crucial for solving the problem effectively.
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Homework Statement


The handle of a floor mop makes an angle θ = 39.6° with the horizontal. Assume the handle is massless, and the mop has mass M = 3.07 kg. The coefficient of kinetic friction between the mop and the floor is μk = 0.386. Find F, the magnitude of the force, exerted downward along the handle, that will cause the mop to slide across the floor without acceleration.


Homework Equations


Frictional force = μ*Normal force
Components
F=ma


The Attempt at a Solution


The bigger problem that I'm not understanding is how to solve the problem. They are asking for a force to push down the handle so that acceleration will become 0, or so that the mop moves at a constant velocity. I don't know where to start...
 
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Remember
Force of gravity at the angle = mgsin(theta)
Force of friction = mgcos(theta) * coefficient of friction
 
MagicRub said:
Remember
Force of gravity at the angle = mgsin(theta)
Force of friction = mgcos(theta) * coefficient of friction

Hi MagicRub,

Thanks for your reply.

Although I do remember those, I don't see how it applies. I thought the mop itself is on the floor, which would be a horizontal surface, and the handle is the part making an angle. The force of gravity would actually be the weight, or mg since it is not on an incline.

Now that I think about it, are you saying to draw the situation, but then rotate it in a way such that the handle of the mop becomes horizontal, and then the mop itself is now at an angle?
 
MagicRub said:
Remember
Force of gravity at the angle = mgsin(theta)
Force of friction = mgcos(theta) * coefficient of friction

How can the force of friction be equal to a force parallel with the handle? Isn't the force of friction between the mop and the floor, which is horizontal? I thought the force of friction would be μs*mg = 11.625

I still don't understand how pushing the handle down would prevent the mop from accelerating because it will always be in contact with the floor..
 
iJamJL said:
How can the force of friction be equal to a force parallel with the handle? Isn't the force of friction between the mop and the floor, which is horizontal? I thought the force of friction would be μs*mg = 11.625

I still don't understand how pushing the handle down would prevent the mop from accelerating because it will always be in contact with the floor..

Since you are pushing down on the mop, the normal force would increase. Remember that the normal force is not defined as the force of gravity. In this case, it would be the sum of the gravitational force and the force that you are pushing downward on the mop with. (keyword:downward; you are pushing at an angle so you need that component)

I believe that MagicRub's equations are for an inclined plane.
 
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