piston 1 has a diameter of 0.25 in; pistol 2 has a diameter of 1.5 in. In the absence of friction, determine the force necessary to support the 500 lb weight.

Answer 1

I will have to guess what your problem is because if you tried to copy and paste a picture it did not work.
If we are talking a hydraulic problem with force on the small piston lifting a weight on the big one then
F =(little area/big area) weight
F = (.25^2/1.5^2) 500
F = [ 1/(4*2.25) ] 500
F = 55.56 lb