A rectangular block 3m high and 1.8m wide weighing in at 7600N is pushed with a force 20.5° to the height of the block 1.6m from the bottom of the block.
Calculate the Force Required to tip over this block.
Torque = perpendicular force x distance or force x perpendicular distance
The Attempt at a Solution
Firstly, I make a free body diagram of the block.
N force up
mg force down
F 69.5° [N o E]
Secondly, I write my ƩTorque eq, using the pivot point where the box will tip on an angle. I will call this P.
Ʃτp = 0
(0.9)mg = (Fsin69.5)(1.6)
F = 4564.0 N