Value of friction with constant velocity

AI Thread Summary
In the discussion about a block being pulled across a rough table at a constant velocity, the focus is on determining the equation for the magnitude of the kinetic friction force, fk. The initial answer provided was μk(mg - T*cos(θ)), but confusion arose regarding the correct formulation. It was clarified that the vertical forces include the y component of the pulling force, leading to the realization that the correct expression should be uk(mg - T*sin(θ)). Participants emphasized that the problem does not require the coefficient of friction, but rather the magnitude of the kinetic friction force itself. The conversation highlights the importance of considering all forces acting on the block to arrive at the correct solution.
jcruise322
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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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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?
 
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.
 
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.
 

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