A solid cylinder (radius = 0.190 m, height = 0.190 m) has a mass of 21.4 kg. Thi

Question

A solid cylinder (radius = 0.190 m, height = 0.190 m) has a mass of 21.4 kg. This cylinder is floating in water. Then oil ( = 867 kg/m3) is poured on top of the water until the situation shown in the drawing results. How much of the height of the cylinder is in oil

A suitcase (mass m = 20 kg) is resting on the floor of an elevator. The part of the suitcase in contact with the floor measures 0.50 m × 0.15 m. The elevator is moving upward with an acceleration of magnitude 1.20 m/s2. What pressure (in excess of atmospheric pressure) is applied to the floor beneath the suitcase?

Explanation / Answer

There is a bouyant force due to the oil, and another buoyant force due to the water. Their sum is equal to the weight of the cylinder.

Bo + Bw = W
po*g*Vo + pw*g*Vw = pc*g*Vc
po*Vo + pw*Vw = pc*Vc
po*(Vo/Vc) + pw*(Vw/Vc) = pc

where:
Vo, Vw, Vc = volume of cylinder in oil, in water, total volume of cyl
po, pw, pc = density of oil, water, cylinder

Further, the portions in oil and water make up the whole, so we have

(Vo/Vc) + (Vw/Vc) = 1

pw =1000kg/m^3

po = 896 kg/m^3

mc = 21.4 kgb

r = 0.19 m , h 0.19m

V =pi*r2h =(3.14*0.19*0.19*0.19) = 0.02154

pc = m/V = 21.4 / 0.02154

pc = 993.63 kg/m^3

(Vo/Vc) = (pc - pw)/(po - pw)= (993.63 -1000)/(896-1000) = 0.06125

The heighog the cylinder is in the oil is 0.06125* 0.19 = 0.01164 m

m = 20 kg

A =0.5mx0.15m = 0.25 m^2

a =1.2 m/s^2

From newton seoond law net force along vertical direction

T -mg = ma

T = m(g+a) = 20(9.8+1.2)

T =220 N

P =Force /ara = 220/ (0.5*0.15)

P = 2933.33 N/m^2

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