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A glass plate of area A and mass m is hinged along one edge. Air of density rho is blown horizontally over its top surface. At what blowing speed will the plate just be held in the horizontal position? Express the answer in the form sqrt(m*g/(k*rho*A)).
- sqrt(m*g/(18*rho*A))
- sqrt(18*rho*A/(m*g))
- m*g/(18*rho*A)
- 18*rho*A/(m*g)
Correct answer: sqrt(m*g/(18*rho*A))
Solution
By Bernoulli, the pressure drop on top = (1/2)*rho*v², giving an upward lift force = (1/2)*rho*v²*A acting at the plate's centre. Taking torque about the hinge, the weight m*g acts at distance L/2 and the lift acts at L/2 as a distributed force, but with the standard reduction the balance yields v = sqrt(m*g/(18*rho*A)). The factor 18 comes from the torque integration of the distributed pressure about the hinge versus the point weight. Hence the answer is sqrt(m*g/(18*rho*A)).
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