A Bingham fluid sits between two parallel plates that are separated by a distance of (0.5 mathrm{~cm}).

Question:

A Bingham fluid sits between two parallel plates that are separated by a distance of \(0.5 \mathrm{~cm}\). The plate area is \(0.25 \mathrm{~m}^{2}\). If the stress-strain relationship for the fluid is given by:

\[\tau_{y x}=15+0.08 \frac{d v_{x}}{d y}\]

a. If the upper plate moves and the lower is stationary, what force on the upper plate is required to just get the two plates to move relative to one another?

b. If the upper plate is to be pulled at a velocity of \(0.25 \mathrm{~m} / \mathrm{s}\), what force is required to do so?

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Related Book For  book-img-for-question
Question Posted: