Question

In the figure, the top tank, which is open to the atmosphere, contains water and the bottom tank contains oil covered by a piston. The tank on the right has a freely movable partition that keeps the oil and water separate. The partition is a vertical distance 0.10 m below the open surface of the water. If the piston in the bottom tank is 0.50 m below the open surface of the water and has a surface area of 8.3 x 10-3 m2, what must the mass of the piston be to keep the system in mechanical equilibrium. For simplicity, ignore the mass of the partition.

Answer 1

okay, If we are to draw a free body diagram for piston, there will be three forces,

force due to air above the piston (F1) ( this force acts downwards)

force due to oil below the piston (F2) ( this force is upwards)

weight force of piston (W) (acting downwards)

we also know that in mechanical equilibrium, acceleration is zero.

so,

using Newton's second law

F_net = ma

F2 - F1 - W = ma

F2 - F1 - W = 0 ------ (1)

we also know

F = P * A where P is pressure and A is cross sectional area

solving (1) for mass of piston

F2 - F1-  mg = 0

P2 * A - P1 * A - mg = 0

m = ( P2 - P1)A/ g -------------- (2)

where P2 is pressure due to oil and can be written as

P2 = P_water + _oilgh2

P2 = P1 + _watergh1 + _oilgh2 ------- (3)

where h1 = 0.10 m and h2 = 0.5 - 0.1 = 0.4 m

just put (3) in (2)

m = ( P1 + _watergh1 + _oilgh2 - P1) A / g

m =  (_watergh1 + _oilgh2 ) A / g

(NOTE - PLEASE CHECK DENSITY OF OIL, THERE ARE SO MANY TYPES OF OIL )

m = ( 1000 * 9.8 * 0.1 + 9200 * 9.8 * 0.4) 8.3e-3 / 9.8

m = 3.8844 Kg