A cubical box is filled with sand and weighs 510 N. It is desired to "roll" the box by pushing horizontally on one of the upper edges. What minimum force (in newtons) is required? Any help is appreciated. Been at this for 30 minutes and I still don't get it.



Answer :

Clockwise moments = Anticlockwise moments
Call the force you apply horizontally to the top left side of the square N.
W = 510 acting from the centre as the mass is distributed evenly.

Take moments about a pivot, in this case the bottom right corner is the pivot.
The clockwise moment is N and the anticlockwise moment is W.
N*x*sqrt(2) = 510*x/2
205x =Nx*sqrt(2)
205 = N*sqrt(2)
N = 205/sqrt(2)

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