What are the forces acting on a sliding box?

[SelenaT]

I was asked to draw a free body diagram to derive the equation tan θ = µs for the following situation:
A box sliding down an inclined plane.

My FBD has 3 forces
- gravitational forces pointing downward (horizontally) <-- longest vector
- normal force perpendicular to the box
- frictional force pointing in the opposite direction of the boxes movement

normal force and frictional force have vectors of the same length.
There is no kinetic or static force right? I'm not missing any forces?

 

[RYANDTRAVERS]

Firstly, what does the 's' represent? Because, otherwise you can just equate the component of the weight along the inclined plane with the normal reaction force, which should give you:
\begin{equation}
\tan \theta = \mu
\end{equation}

 

[PhanthomJay]

I was asked to draw a free body diagram to derive the equation tan theta = µs for the following situation:
A box sliding down an inclined plane.

If you are looking for the static friction coefficient , the box cannot be sliding down the incline. It is just on the verge of about to slide at the angle of incline theta

My FBD has 3 forces
- gravitational forces pointing downward (horizontally) <-- longest vector

gravity force always points down ...it has components down the plane and perpendicular to the plane

- normal force perpendicular to the box
- frictional force pointing in the opposite direction of the boxes movement

yes

normal force and frictional force have vectors of the same length.
There is no kinetic or static force right? I'm not missing any forces?

thats all the forces, but why should the normal and friction forces be equal in magnitude? And why do you say there is no static (friction ) force? Draw free body diagram and apply Newton1st law in the directions down and perpendicular to the plane.