Torques acting on a decelerating box - Maximum height of the box so it does not tip over [closed]

A square based box (length and width = 1m) lies on a horizontal surface on a train. The train decelerates constantly at 5ms^-2 till it stops. Determine the maximum height of the box so that it does not tip over.

Assumptions: - Box does not slide - Equal weight distribution

My answer: the 2 forces providing rotational torque are 1. weight of box acting on the centre of mass and 2. the horizontal deceleration of the object (5ms^-2) also acting on the centre of mass. Taking moments at the pivot point - corner of the box:

Torque of 1. = 9.8M * 1/2 (M=mass of box)

Torque of 2. = -5M * h/2 (h= height of the box)

For the box not to tip:

1.+ 2. = 0

9.8M * 1/2 = 5M * h/2

9.8 = 5 *h

h= 9.8/5 = 1.96 m

Very simple calculation so I am worried it is incorrect...

Thanks in advance for the help.

closed as off-topic by John Rennie, Martin, John Duffield, ACuriousMind♦, WolpertingerAug 27 '16 at 17:53

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