A matchbox of height h and uniform density stands on its end on a table with coefficient of friction μ. For accelerating it horizontally, at acceleration a, whilst upright, at which height y should the acceleration be applied?

Since the matchbox is of uniform density, the centre of mass is at the centre. In order to accelerate it horizontally, it must have zero resultant moment about the centre of mass, else it will rotate and topple.The matchbox will experience a moment of rotation from the pushing force at a distance (h/2 - y) from the centre of mass, and a moment of rotation in the opposite direction from the frictional force, at a distance h/2 from the centre of mass.The equation of motion for the pushing force in the horizontal direction is:Fpush - μmg = masince the resistance force from friction will be μ*R, where R is the reaction force and will in this case equal mg as the matchbox stands upright.Rearranging this gives Fpush = m(μg + a)
Taking moments about the centre of mass of the matchbox,(h/2-y)m(μg+a) = h/2μmg
Solving this equation givesy = (ha)/2(μg+a)