A student could either push or pull, at an angle of 30 degrees from the horizontal, a 40kg crate, where the coefficient of kinetic friction is .21. The crate is moved 18m. Calculate the minimum work for pushing and pulling.
W=F•(change in)X•cos(angle in between the force and direction of motion)
The Attempt at a Solution
So, it seems like this problem has to be solved with systems of equations. Since I do not know the force in Newtons itself, I went about creating a free body diagram to find what the forces should be.
With my attempt, I did an X-analysis.
For the first FBD, the force has an angle of 30 degrees above the horizontal.
F=ma (acceleration is 0, I am assuming, as the force would be constant).
F - f = 0 [magnitude of the force in the X direction minus kinetic friction]
Fcos(30) - u•N = 0 [force in X direction minus the coefficient of friction times normal—this is where things get shady]
F = ((u•N)/cos(30)) [total force is going to equal the mue times normal force divided by cosine of 30 degrees]
~ but what is N? simply, it is -(m•g), but let's go to the y-analysis to double check, right? ~
F=ma (acceleration is 0)
N + F - W = 0 [the normal force plus the y component of the force minus weight is zero)
N + Fsin(30) - mg = 0
N = -Fsin(30) + mg
So, we have the normal force equation. But, if we plug this in for N on the X-analysis, it just ends up cancelling the force. Same goes for if you try to create a system of equations for F. I do understand that for systems of equations, the one you are plugging into needs to be equal to zero. I did this..., and it still did not work out. Am I just not doing my system of equations right? Or is normal more simple than this?
I tried the SoEs many times, but I can't find my attempts right now so I won't post them.., can somebody help me out? Now I am all messed up and can not move on until I figure this one out!
Help! Oh, and thanks :)
P.S.: do you only need to find the normal force in equations where the normal force is at a different angle?