Answer
To tip the crate, the force must be applied at a minimum height of $~~0.536~m~~$ above the floor.
Work Step by Step
We can consider the torque about a rotation axis located at the position of the small obstruction.
There is a torque about this rotation axis from the weight of the crate. To tip the crate, there must be an opposing torque from the applied force which overcomes the torque from the weight of the crate.
Let $h$ be the height of the applied force. We can find the minimum height $h$:
$\tau_{net} = 0$
$h~(350~N)-(0.375~m)(500~N) = 0$
$h~(350~N)=(0.375~m)(500~N)$
$h = \frac{(0.375~m)(500~N)}{350~N}$
$h = 0.536~m$
To tip the crate, the force must be applied at a minimum height of $~~0.536~m~~$ above the floor.
