Question

an object Of weight W has auniform rectangal an object of weight W has a uniform and rectangular an object of weight W has a uniform rectangular area of cross section of a into a and density of 0.25 gram per centimetre cube part of it is immersed in water and the rectangle is tilted by 45 degree while one of its corners is just at the water surface find the torque of the buoyancy
force with respective centre of mass of the object ​

Answer 1

The value of torque of bouncy is τ = aω / 2 √2  

Explanation:

Given data:

Angle of tilt = 45°

Density of area = 0.25 g/cm^3

d = a/2 sin 45°

d = a / 2√2

Fb = 1/2 x a√2  x a/√2 x 1

Fb = a^2g/2

τ = Fb x d

  =  a^2g/2 x a / 2√2  

  = a^3g/4 √2                      

ω =  0.25 x 2a x a x 1 x g

ω =  a^2g/2

Now

τ = aω / 2 √2  

Thus the value of torque of bouncy is τ = aω / 2 √2  

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