Calculating Normal Force on a Block in Equilibrium

[StephenDoty]

If a block of mass m is pulled in the direction of (x-component) Tcos(theta) and (y-component) Tsin(theta) across a rough surface with a constant velocity. What is the magnitude of the frictional force?

A) μkmg B) μkT cos θ C) μk(T – mg) D) μkT sin θ E) μk(mg – T sin θ)

Since fk is in the opposite direction as the motion and since there is no acceleration F=0=Tcos(theta) - fk
So
fk= Tcos(theta)

So is the answer B?

 

[Doc Al]

clarification

If a block of mass m is pulled in the direction of (x-component) Tcos(theta) and (y-component) Tsin(theta) across a rough surface with a constant velocity. What is the magnitude of the frictional force?

Is the x-y plane parallel to the surface? (I would assume yes.) Or is the y-direction vertical?

 

[Littlepig]

Edit: (supposing x is horizontal+ to right and y is vertical+ to top)

So, you know that friccional force is proporcional to Normal force exerced to the block;

you know that if dv/dt=0 => Sum of all Forces=0, so,
begin with making a diagram of the problem, putting every vector force on it with magnitudes that correspond to total F=0...
than you need to understand why isn't B)

 

[StephenDoty]

y is vertical and x is horizontal
If the fnet = 0 then
0=Tcos(theta) - fk
which makes fk= Tcos(theta)

If it is not B then I do not understand how to solve this problem.

 

[Doc Al]

y is vertical and x is horizontal
If the fnet = 0 then
0=Tcos(theta) - fk
which makes fk= Tcos(theta)

This is true, but it's not one of the choices. But there is another correct choice.

 

[StephenDoty]

I am at a loss
I know that fk usually equals ukmg on a horizontal force. Any help would be appreciated.

 

[Doc Al]

fk = ukN. But does N always equal mg?

 

[StephenDoty]

If N= Tsin(theta)
then Fk= uTsin(theta)

Answer D

 

[Doc Al]

If N= Tsin(theta)

Figure out the normal force by considering the vertical forces acting on the block.

 

[StephenDoty]

ok
the normal force is the y-component and the force in the opposite direction is the force of gravity on the mass
so 0=mg-Tsin(theta)
so mg=Tsin(theta) so N=Tsin(theta)
so fk = uk (Tsin(theta))

what am I missing?

 

[Doc Al]

There are three vertical forces acting on the block: The vertical component of T, the weight, and the normal force. They must add to zero. Solve for N.