# Answer to Question #26121 in Mechanics | Relativity for cesarpintor

Question #26121

A solid cylinder of wood has a cross sectional area of 20cm2 and height of 90cm. It floats upright in water. If two thirds of the length of the cylinder is submerged, calculate: the weight of the water displaced?

Expert's answer

According to the condition, volume of displaced waterequals 2/3 of cylinder's volume. We can easily find cylinder's volume:

V=S*h,

S - cross sectional area; h-height of the cylinderthen volume of displaced water is V1=2/3*S*h; The weight of the displaced water

volume V1 is P=ρgV1, where ρ - density of water, g - gravitational acceleration.

So, P=2/3*ρgSh; P=2/3*1000*10*20*10^(-4)*0.9=12 (N)

V=S*h,

S - cross sectional area; h-height of the cylinderthen volume of displaced water is V1=2/3*S*h; The weight of the displaced water

volume V1 is P=ρgV1, where ρ - density of water, g - gravitational acceleration.

So, P=2/3*ρgSh; P=2/3*1000*10*20*10^(-4)*0.9=12 (N)

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