Question 5.13: A vane is attached to a block as shown in Fig. 5.17. The blo...

Choose the CV and coordinate systems as shown in Fig. 5.17(a). X Y is the inertial reference frame, while reference frame xy moves with the cart.

Applying x component of linear momentum conservation equation to the linearly accelerating control volume, we get

-\mu_{k} M_{b} g-M_{b} \frac{d U_{b}}{d t}=0+\rho A\left(V_{j}-U_{b}\right)\left[-A\left(V_{j}-U_{b}\right)\right]                     (5.38)

It is important to note here that the CV constitutes of a solid portion (block) and a fluid portion (liquid). The fractional occupancy of the liquid in the CV is small enough so that its contribution to the velocity relative to the xy reference frame (i.e., V_{xy}) may be practically neglected. Thus, the first term on the right-hand side of the momentum equation turns out to be approximately zero. In addition, the exit of water relative to CV has orientations only along the vertical (one stream upward and another stream downward), leaving no outward flux of momentum along the x direction. Accordingly, one may obtain the linear momentum equation ( Eq. (5.38)) in the following simplified form:

M_{b} \frac{d U_{b}}{d t}-\rho A\left(V_{j}-U_{b}\right)^{2}+\mu_{k} M_{b} g=0                               (5.39)

Equation (5.39) may be rearranged as

With an initial condition:

At terminal speed,  \frac{d U_{b}}{d t}=0  and  U_{b}=U_{b t} ,  so

Or                      V_{j}-U_{b t}=\sqrt{\frac{\mu_{k} M_{b} g}{\rho A}}

Or                  U_{b t}=V_{j}-\sqrt{\frac{\mu_{k} M_{b} g}{\rho A}}                      (5.40)