Value of friction with constant velocity

[jcruise322]

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?

 

[SteamKing]

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?

 

[jcruise322]

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.

 

[SammyS]

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.

μ_k does not need to be used here.