A pump is partially submerged in oil and is supported by 4 springs. The oil has a specific gravity of 0.90, the weight of the pump is 14.6 lb, and the submerged volume is 40 in3. What force is needed in the springs so that the pump will remain in static equilibrium

submerged volume of pump = 40 x 12⁻³ ft³

density of oil =.9 x 62.4 lb ft⁻³ (specific gravity x density of water)

mass of displaced oil = 40 x 12⁻³ x. 9 x 62.4 lb

= 1.3 lb

buoyant force = 1.3 lb

weight of pump = 14.6 lb

extra force needed for balancing pump by spring

= 14.6 - 1.3

= 13.3 lb.