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Mechanical Properties Of Fluids

Question
CBSEENPH11020016

A plane is in level flight at constant speed and each of its two wings has an area of 25 m2. If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m–3).

Solution
Given,
Area of the wings of the plane, A = 2 × 25 = 50 m2

Speed of air over the lower wing, V1 = 180 km/h = 50 m/s 
Speed of air over the upper wing, V2 = 234 km/h = 65 m/s
Density of air, ρ = 1 kg m–3

Pressure of air over the lower wing = P1

Pressure of air over the upper wing= P2

The upward force on the plane can be obtained using Bernoulli’s equation,

space space space space space space space space space straight P subscript 1 space plus space 1 half ρV subscript 1 squared space equals space straight P subscript 2 space plus space 1 half ρV subscript 2 squared

rightwards double arrow space space space space straight P subscript 1 space minus space space straight P subscript 2 space space equals space 1 half straight rho open parentheses straight V subscript 2 squared minus straight V subscript 1 squared space close parentheses space space space space space space space space space space space... space left parenthesis straight i right parenthesis

Upward space force space straight F space on space the space plate space is comma space
fraction numerator straight P subscript 1 space minus space space straight P subscript 2 over denominator straight A end fraction space equals space 1 half space space space space space space space space space straight P subscript 1 space plus space 1 half ρV subscript 1 squared space equals space straight P subscript 2 space plus space 1 half ρV subscript 2 squared

rightwards double arrow space space space space straight P subscript 1 space minus space space straight P subscript 2 space space equals space 1 half straight rho open parentheses straight V subscript 2 squared minus straight V subscript 1 squared space close parentheses space space space space space space space space space space space... space left parenthesis straight i right parenthesis

Upward space force space straight F space on space the space plate space is comma space
fraction numerator straight P subscript 1 space minus space space straight P subscript 2 over denominator straight A end fraction space equals space 1 half space space space space space space space space space straight P subscript 1 space plus space 1 half ρV subscript 1 squared space equals space straight P subscript 2 space plus space 1 half ρV subscript 2 squared

rightwards double arrow space space space space straight P subscript 1 space minus space space straight P subscript 2 space space equals space 1 half straight rho open parentheses straight V subscript 2 squared minus straight V subscript 1 squared space close parentheses space space space space space space space space space space space... space left parenthesis straight i right parenthesis

Upward space force space straight F space on space the space plate space is comma space
space space left parenthesis straight P subscript 1 space minus space space straight P subscript 2 right parenthesis thin space straight A space equals space 1 half space straight rho space open parentheses straight V subscript 2 squared minus straight V subscript 1 squared space close parentheses space straight A

space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 half cross times 1 cross times open parentheses 65 space squared minus 50 squared space close parentheses space cross times 50 space

space space space space space space space space space space space space space space space space space space space space space space space space equals space 43125 space straight N space

Using space Newton apostrophe straight s space first space law comma space mass space of space the space plane space can space be space obtained. space

straight F space equals space mg

therefore space straight m space equals space fraction numerator 43125 over denominator 9.8 end fraction space equals space 4400.51 space kg

Therefore comma space the space mass space of space the space plane space is space
approximately space 4400 space kg.