I suspect the difference in gravity between when the Moon is overhead (or at the nadir) and when it is sideways would be much less than the effect of the constant small tremors that would be felt if the object was set on solid ground.
Maybe a kind of low-pass system would allow it to react to slower change and ignore the higher amplitude, higher frequency of the tremors， though.
Probably not. The moon has a mass of about 7.34e22 kg, the minimum distance from the moon to the surface of the earth is about 363,104 km, or 363,104,000 meters, and the constant of proportionality for gravity is G = 6.673e-11 N m^2/kg^2, so the attraction between the Earth and a 1kg object on the surface of the Earth is 3.7e-5, or about 8.3e-6 pounds force. That's not much to work with.
It's more complex than that. The floor an object sit's on is also attracted to the moon so you would need to look at the difference in attraction over the size of the device.
However, the earths crust deforms a few cm due to tidal forces. Which weekens local gravity. Aparently you can measure tides in a trees sap so detecting them on a small scale is probably possible. (My google fo is failing to find that one though.)
It is certainly more complex than that, but I think that a back-of-the-envelope calculation like the one above is more than enough to strongly suggest that getting a solid object to remain in place when the moon is above it and not when the moon is below is very unlikely to be feasible.
