Work & Energy: Forces with Angles

[LiterallyLogan]

Homework Statement

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.

Homework Equations

W=F•(change in)X•cos(angle in between the force and direction of motion)
F(net)=ma

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.
Thus:
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? ~

Y-analysis.
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?

 

[Doc Al]

~ but what is N? simply, it is -(m•g), but let's go to the y-analysis to double check, right? ~

As your analysis shows, the normal force does not equal 'mg'. (It would if there were no vertical component of F.)

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.

Why do you think it cancels? Show how you plugged it in.

 

[LiterallyLogan]

OK, after coming back to the problem, I think I have figured it out:

So, X-analysis;
Fx - f = 0
Fcos30 - f = 0
Fcos30 = f
F = (u•N)/(cos30) [this is the equation for force that we can plug in]

Y-analysis;
N - W + Fsin30 = 0
N - W + ((u•N)/cos30)•sin30 = 0
N - W + (u•N)•tan30 = 0
N + (u•N)•tan30 = W
N + N = (W)/(u•tan30)
2N = W/(u•tan30)
N = (1/2)(W/(u•tan30))
N = (1/2)(mg/(u•tan30))
Thus,
N = .5(40kg•9.8(m/s^2))/(.21•tan30)
N = 1616.58[kg•(m/s^2)] = 1616.58N.

Algebra has never been my strong suit; I have taken calculus-based physics, which seemed a bit easier to me than algebra based physics that does not allow graphing calculators.

 

[LiterallyLogan]

To find the second N, I have done this, but I do not feel as confident:

X-Analysis:
Fx - f = 0
Fsin30 - f = 0
Fsin30 = f
F = (f)/(sin30) [equation to plug in for F]

Y-Analysis:
N - W - Fy = 0
N - W - Fcos30 = 0
N - W - ((f)/sin30)•cos30
N - W - ((f)/tan30)
N - ((uN)/tan30) = W
N - ... at this point, I don't know what to do, as tan30 is not part of both sides of the left side of the equation, so I can not transfer it.
Looking at the top part as well of my previous work, I am pretty sure that is not correct as I made the same mistake above, like here:

N - W + (u•N)•tan30 = 0
N + (u•N)•tan30 = W
N + N = (W)/(u•tan30)

I don't believe that is correct, as to multiple tan30 to get it to the other side (or divide, my brain is dead at this point), both sides of the left side would have to contain tan30, correct? *Sigh*...

 

[Doc Al]

OK, after coming back to the problem, I think I have figured it out:

So, X-analysis;
Fx - f = 0
Fcos30 - f = 0
Fcos30 = f
F = (u•N)/(cos30) [this is the equation for force that we can plug in]

That's fine.

Y-analysis;
N - W + Fsin30 = 0
N - W + ((u•N)/cos30)•sin30 = 0
N - W + (u•N)•tan30 = 0
N + (u•N)•tan30 = W

So far, so good.

N + N = (W)/(u•tan30)

This step is an error. Redo it.

My suggestion: Write your two equations for the x and y analysis. But instead of solving for N, solve for F. That's what you need to solve for the work done.