# Minimum force required to turn over box.

1. Jan 1, 2014

### peripatein

Hi,
1. The problem statement, all variables and given/known data
I would like to determine the position, magnitude and direction of the minimum force F required to turn over the box shown in the attachment (width - B, height - H). The (uniform) box is placed on a horizontal plane. The coefficient of friction between the box and the plane is μ.

2. Relevant equations

3. The attempt at a solution
Since τ=r x F, I believe the minimum force would have to be at either D or C (so that r is maximum). I picked C and placed my Fx and Fy there, where Fx=Fcosθ and Fy=Fsinθ and θ is the angle between F and the positive x axis. I was then trying to write the equations for the torques wrt to A. I figured that when the box is turned over, N equals zero and therefore there is no friction. I was then left with two effective forces, namely: F and mg.
The equation I wrote down was:
BFsinθ - FHcosθ-mgB/2=0
Is that correct?

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2. Jan 2, 2014

### tiny-tim

hi peripatein! happy new year!
yes
(how can N = 0 ?? )

no, the criterion for any object to start to tip is that N goes through the edge of its base (in this case, A or B)

do ∑moments = 0 to find when that happens

(technically, this assumes that the box is not sliding

if it does slide, you will also need to consider the acceleration)​