Solving a Grade 12 Physics Problem: Pulling vs Pushing a Heavy Box

[JimmyRay]

Bonus Question...

So my teacher gave us this question... where he describes that. "We have a heavy box, we want to move it but can NOT apply horizantal force (meaning CANT get on your knees and push it) we would have to either push down on it at an angle to move it, or pull up and push (or pull he didnt specify), which method is easier?"

Ok well I would say that pulling up on it would be easier because it reduces the amount of friction, right? Pushing down on it, increases the friction... But the hard part is...

Our answer has to be in the form of an algebraic equation..

How do I go about converting my answer into algebra (assuming my answer was right).

He said it's a grade 12 problem, but we should be able to do it... (using our knowledge of physics, grade 11 formulas ..)

Ive seen some grade 12 problems... the only difference right now for me between grade 11 and 12 problems is that they throw in some trig... Would I have to use trig here because you're pulling up, the surface isn't horizantal? Or could I use grade 11 knowledge and make an equation?

 

[HallsofIvy]

Draw two pictures, one showing a force pushing down on the box and the other showing a force pushing up. Yes, you will need to use trigonometry to divide the force into horizontal and vertical components. In one the vertical component will be downward, adding to the weight and increasing the friction. In the other the vertical component will be upward, reducing the weight and decreasing the friction.

 

[JimmyRay]

I drew two diagrams, the thing is that I understand what you're saying... but I still can't express it as an equation...
im trying to come up with an equation for the first FBD I drew which has the downward force on it.

Y Dir-

Fn(normal force) - Fd (force downward) - Fg (force of gravity) = 0

...Thats all I have so far... :(

 

[JimmyRay]

Wait wait wait wait

... I tried again and for my first FBD (for the downward force) I got

Ff = MFn
Fn = mg + Fdsin
Ff = M(mg+Fdsin)

And that would mean that for the second one it would be

Ff = M(mg-Fusin)


...

Can I go further?