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A force of 10 N is needed to drag a rectangular glass plate across a liquid surface at a certain speed. What force is required to drag, at the same speed, a second glass plate that is 3 times as long and 2 times as wide?
- 5/3 N
- 10 N
- 60 N
- 30 N
Correct answer: 60 N
Solution
The viscous force F = eta*A*(dv/dz). At the same speed and gap, F is proportional to the wetted area A = length * width. The new area is (3*length)*(2*width) = 6 times the original. So the force becomes 6*10 = 60 N.
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