Frame on castors toppling over

[D19A99G]

Hello all,
I'm trying to calculate how much force is required to topple over a frame with a conveyor attached to it. I have attached a screenshot of the assembly.
I have the centre of gravity as seen on the screenshot. The two lines to the pivot points show the angle. These are 16 degrees (line length of 1032mm) and 12.5 degrees (line length of 1015mm).
The total mass of the unit is 175kg.

I have seen a couple of methods online such as using kinetic and potential energy or trying to calculate the torque required to push the CoG over the pivot.

I'm struggling with this seemingly simple task! Any input would be appreciated, if anymore information is required let me know.

Thanks,

 

[Baluncore]

Welcome to PF.
Mass is 175 kg. Vertical force = m·g = 175 * 9.8 = 1715 Newton.
Push on the 16° side, fall towards 12.5° side. 1715 * Tan( 12.5°) = 380.2 N
Push on the 12.5° side, fall towards 16° side. 1715 * Tan( 16.0°) = 491.8 N

Edit: The force is assumed to be applied to the centre of mass.

 

[Baluncore]

Height of the centre of mass is; 1.015 m * Cos( 12.5° ) = 0.992 m
Check height of centre of mass; 1.032 m * Cos( 16.0° ) = 0.992 m

Push on the 16° side, fall towards 12.5° side. 1715 * Tan( 12.5°) = 380.2 N.
Which represents a torque of 380.2 N * 0.992 m = 377.2 N·m.
If the force is applied at height of h metre, the force in Newtons will be; F = 377.2 / h Newton.

Push on the 12.5° side, fall towards 16° side. 1715 * Tan( 16.0°) = 491.8 N.
Which represents a torque of; 491.8 N * 0.992 m = 487.9 N·m.
If the force is applied at height of h metre, the force will be; F = 487.9 / h Newton.

 

[D19A99G]

Thank you for the response!
How would the swivel castors then be factored into that? Is there any easy way?

 

[Baluncore]

One failure is one industrial accident too many. The easiest way to factor in the swivel castors is to assume the castor will sometimes be blocked from turning or rolling by a small object on the floor. That happens more often than you might expect.