Question

An iron cylinder and a silver cylinder have the same length, L = 10cm, and different cross-sectional areas, ARe = 4 cm 2 , and AAg = 5.0cm2 . The cylinders are connected to each other by a string which passes over a pulley. Initially they are held in place so that each is half submerged in mercury. When the cylinders are released, the iron cylinder moves upward through a distance x and the silver cylinder moves downward through a distance x, allowing the system to come to equilibrium. Find x. You will need the densities of iron, silver, and mercury, which can be found in your textbook.

Answer 1

x=1.1644 cm

Step-by-step explanation:

The first step is to make a sum of forces in each cylinder:

1) Iron cylinder: In this cylinder there are three forces, the buoyant force(Fbi) more the force of the silver cylinder (Fs) transmitted through the polley less the iron weight (Wi), so:

`Fbi+Fs=Wi`

`Fs=Wi-Fbi`

2) Silver cylinder: In this cylinder there are three forces too, the buoyant force(Fbs) more the force of the iron cylinder (Fi) transmitted through the pulley less the silver weight (Ws), so:

`Fbs+Fi=Ws`

`Fi=Ws-Fbs`

Now, we have a simple pulley, by the principle of the pulleys we know that the force of the silver cylinder must be the same as the force of the iron cylinder, then:

`Fs=Fi`

`Wi-Fbi=Ws-Fbs`

To calculate the buoyant forces we have to consider the submerged volumen of each cylinder, from the problem we know that the iron cylinder moves upward "x" distance from the center of the cylinder thus the iron submerged volumen (Vis) is:

`Vis= (5-x)(4) cm3`

And the silver cylinder moves downward "x" distance from the center of the cylinder thus the silver submerged volumen (Vss) is:

`Vss= (5+x)(5) cm3`

The densities of the iron, silver and mercury are known:

ρiron=0.00784 kg/cm3

ρsilver=0.01047 kg/cm3

ρmercury=0.01047 kg/cm3

Substituting all known values in the obtained equation:

`Wi-Fbi=Ws-Fbs`

(ρiron)(Viron)(g)-(ρmercury)(Vis)(g)=(ρsilver)(Vsilver)(g)-(ρmercury)(Vss)(g)

We can cancel the gravity:

(ρiron)(Viron)-(ρmercury)(Vis)=(ρsilver)(Vsilver)-(ρmercury)(Vss)

Substituting:

`(0.00784)((4)(10))-(0.01356)(5-x)(4)=(0.01047)((5)(10))-(0.01356)(5+x)(5)`

Making the operations:

`0.3136-0.2712+0.05424x=0.5235-0.3390-0.0678x`

`0.12204x=0.1421`

`x=(0.1421)/(0.12204)`

[x=1.1644 cm