The minimum distance will be achieved if the force is the maximum possible value without slipping, which is F = M*g*mus = 90.65 N
If the crate tips instead, the application height of the force H is such that there is a torque balance with the weight acting at the center of mass.
M*g*(1.21/2) = F*H
H = M*g*(1.21/2)/(M*g*mus) = 1.21/(2*mus) = 1.06 m
A uniform crate with a mass of 16.2 kg rests on a floor with a coefficient of static friction equal to 0.571. The crate is a uniform cube with sides 1.21 m in length. what is the minimum height where the force F can be applied so that the crate begins to tip before sliding ?
I know that the horizontal force applied to the top of the crate that initiate the tipping is 79.5 N
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