# Value of friction with constant velocity

1. Dec 15, 2015

### jcruise322

1. The problem statement, all variables and given/known data
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))

2. Relevant equations
Fk=Fn*uk
mg*cos(theta)=magnitude of force in x direction

3. 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.

2. Dec 15, 2015

### SteamKing

Staff Emeritus
For a block of mass m and acceleration due to gravity g, what is Fn?

3. Dec 15, 2015

### 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.

4. Dec 15, 2015

### SammyS

Staff Emeritus
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.