Value of friction with constant velocity

In summary, the equation for the magnitude of the kinetic friction force, fk, is μk(mg-T*sin(θ)). This accounts for the vertical and horizontal components of the pull, as well as the acceleration due to gravity and the angle of the pull.
  • #1
jcruise322
36
1

Homework Statement


A block of mass "m" is pulled across a rough table by a string at an angle "θ" above the horizontal at a constant velocity. What is an equation for the magnitude of the kinetic friction force, fk?

The answer that I was given was μk(mg-T*cos(θ))

Homework Equations


Fk=Fn*uk
mg*cos(theta)=magnitude of force in x direction[/B]

The Attempt at a Solution



I just thought it would be mg*cos(theta)=uk*Fn

Since there is no acceleration, the frictional force in the positive x direction must equal the force in the negative x direction...so, T*cos(theta), which is the same in this situation as uk*Fn.
Why that weird answer? Anybody?
 
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  • #2
jcruise322 said:

Homework Statement


A block of mass "m" is pulled across a rough table by a string at an angle "θ" above the horizontal at a constant velocity. What is an equation for the magnitude of the kinetic friction force, fk?

The answer that I was given was μk(mg-T*cos(θ))

Homework Equations


Fk=Fn*uk
mg*cos(theta)=magnitude of force in x direction[/B]

The Attempt at a Solution



I just thought it would be mg*cos(theta)=uk*Fn

Since there is no acceleration, the frictional force in the positive x direction must equal the force in the negative x direction...so, T*cos(theta), which is the same in this situation as uk*Fn.
Why that weird answer? Anybody?
For a block of mass m and acceleration due to gravity g, what is Fn?
 
  • #3
Oh, I see now...there are three vertical forces acting on the object. I forgot about the vertical y component of the pull. Thanks, SteamKing. And, it should be uk(mg-T*sin(theta)). not T*cos(theta). Appreciate it.
 
  • #4
jcruise322 said:
Oh, I see now...there are three vertical forces acting on the object. I forgot about the vertical y component of the pull. Thanks, SteamKing. And, it should be uk(mg-T*sin(theta)). not T*cos(theta). Appreciate it.
You are NOT being asked for the coefficient of friction. You're being asked for the magnitude of the kinetic friction force.

μk does not need to be used here.
 

Related to Value of friction with constant velocity

1. What is friction?

Friction is the resistance force that occurs when two surfaces come into contact and move against each other. It is caused by the microscopic irregularities on the surfaces and can slow down or prevent motion.

2. How does friction affect constant velocity?

Friction has a tendency to slow down objects in motion, so it can decrease the velocity of an object moving at a constant speed. However, if the object is already moving at a constant velocity, the frictional force will be balanced by an equal and opposite force, allowing the object to maintain its constant velocity.

3. Is friction always present with constant velocity?

Yes, friction is always present when two surfaces are in contact and moving against each other. However, the amount of friction may vary depending on the surfaces and the force applied.

4. How can friction be reduced in a system with constant velocity?

Friction can be reduced by using lubricants, such as oil or grease, which create a slippery layer between the surfaces and reduce the amount of contact. Additionally, using smoother surfaces or reducing the force applied can also decrease friction.

5. Can friction ever be beneficial in a system with constant velocity?

Yes, friction can be beneficial in certain situations. For example, in some machines, friction is necessary for the proper functioning of gears and other moving parts. Additionally, friction can also be used to slow down or stop an object in motion, which can be important for safety reasons.

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