How to calculate the right motor size for a soil sifter
If the soil is a solid mass that is stuck to and moves with the tray, that's an upper bound for the moving mass, of about 9kg.

If the soil is a solid mass which slides frictionlessly over the tray, that is a lower bound for the moving mass, of about 2kg.

Unfortunately, in neither case is any work done on the soil, so the power needed is zero, plus losses. 

Maybe that's the wrong tack. The movement of the soil could be a really horrible thing to compute. Let's do the physicist thang and take an upper bound. Let's assume a motor that runs at constant speed, and provides whatever force the shaker system loads it with. Then calculate the maximum instantaneous power as peak force * speed.

The 'stuck to the tray' bound provides an upper bound for force, as when the soil slips, to be 'eaten' by the grid, the force required to accelerate it will be less. So we will go with the 9kg moving mass.

For 6" = 150mm peak to peak motion, r = 0.075m amplitude, at 132 cycles/minute = 2.2Hz, \$\omega\$= 14 rad/second, the approximate equation of motion for a tray driven by a crank is

\$ y = r\sin(\omega t)\$

the speed is \$\dfrac{dy}{dt} = -r\omega\cos(\omega t)\$

and the acceleration is \$\dfrac{d^2y}{dt^2} =  -r\omega^2\sin(\omega t)\$

Substituting back, we get the peak speed to be about 1m/s, and the peak acceleration to be 14m/\$s^2\$. With a moving mass of 9kg, that means a peak accelerating force of 132N, or about 132 watts for the product of peak speed and peak force.

So, with a constant speed motor (which is probably a reasonable assumption for a 'large' motor running at high speed, geared down, so a good energy storage flywheel effect), the power peaks taken from it will be 132 watts, or about 1/5th HP in the (furlong, cwt, fortnight) system of units. The mean power will therefore be less.

Remember this is an upper bound, the actual power required for real soil will be less again. Perhaps if you assumed a solid mass of soil, with a friction coefficient to the tray. Set up the equation of motion for soil and tray as a function of friction. Compute actual power required, then differentiate with respect to friction to find the maximum power. But then again, it might just be easier to use the 1/5th HP motor. Perhaps an air motor would be ideal, from an industrial air-powered nut driver, it might even be geared down to the right speed. A horse-mill would probably be impractical, with the horse feed an all. An electric motor is certainly an option.

Look up 'pumpjack' on wikipedia, to see how the motor used for that application is a high slip type, so that energy can be delivered from the flywheel during the period of high loading.