How does the FBD of this look like?
A furniture crate of mass 56.8 kg is at rest on a loading ramp that makes an angle of 21.1 degrees with the horizontal. The coefficient of kinetic friction between the ramp and the crate is 0.192. What force (in Newtons). applied parallel to the ramp, is required to push the crate up the incline at a constant speed?

Question

How does the FBD of this look like?
A furniture crate of mass 56.8 kg is at rest on a loading ramp that makes an angle of 21.1 degrees with the horizontal. The coefficient of kinetic friction between the ramp and the crate is 0.192. What force (in Newtons). applied parallel to the ramp, is required to push the crate up the incline at a constant speed?

harman.singh · Answer

When the box is in motion, the FBD would look like this
|dw:1478601403656:dw|

harman.singh · Answer

Sorry, forgot to include the frictional force. I will redo it

steve816 · Answer

In what direction will the friction vector be at?

harman.singh · Answer

|dw:1478601676880:dw|

steve816 · Answer

Ok, thanks for the drawing. Another question: If the speed required to push the the crate is constant, would the Fnet = 0 ?

harman.singh · Answer

The question is basically asking for whats the force needed to overcome friction. In that case, you WILL have a net force with a change in acceleration starting from rest. It wont be zero. 

It will only be zero once its in  motion at a constant speed, which occurs later on.

steve816 · Answer

Ah, I see now. Thanks for the response.