Related questions with answers

If a force of P=2N causes the $30$ -mm-diameter shaft to slide along the lubricated bearing with a constant speed of $0.5 \mathrm{~m} / \mathrm{s}$ , determine the viscosity of the lubricant and the constant speed of the shaft when $P=8 \mathrm{~N}$ . Assume the lubricant is a Newtonian fluid and the velocity profile between the shaft and the bearing is line ar. The gap between the bearing and the shaft is $1 \mathrm{~mm}$ .

A thin plate moves between two parallel, horizontal, stationary flat surfaces at a constant velocity of $5 \mathrm{~m} / \mathrm{s}$ . The two stationary surfaces are spaced $4 \mathrm{~cm}$ apart, and the medium between them is filled with oil whose viscosity is $0.9 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^2$ . The part of the plate immersed in oil at any given time is $2$ - $\mathrm{m}$ long and $0.5$ - $\mathrm{m}$ wide. If the plate moves through the mid-plane between the surfaces, find the force required to maintain this motion. What would your response be if the plate was $1 \mathrm{~cm}$ from the bottom surface $\left(h_2\right)$ and $3 \mathrm{~cm}$ from the top surface $\left(h_1\right)$ ?

A $50-cm imes 30-cm imes 20-cm$ block weighing 150 N is to be moved at a constant velocity of 0.80 m/s on an inclined surface with a friction coefficient of 0.27. (a) Determine the force F that needs to be applied in the horizontal direction. (b) If a 0.40-mm-thick oil film with a dynamic viscosity of $0.012 Pa\cdot s$ is applied between the block and inclined surface, determine the percent reduction in the required force.

Question

If the kinematic viscosity of glycerin is $v=1.15\left(10^{-3}\right) \mathrm{m}^2 / \mathrm{s}$ , find its viscosity in FPS units. At the temperature considered, glycerin's specific gravity is $S_g=1.26$

Solution

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First, let us convert the kinematic viscosity of glycerin from SI to FPS units. Since $\nu_g=1.15\cdot 10^{-3}\ \mathrm{m^2/s}$ , we have to convert meters to feet. As we know, it holds that $1\ \mathrm{m}=3.28\ \mathrm{ft}$ . So, $1\ \mathrm{m^2}=10.8\ \mathrm{ft^2}$ , and we obtain:

\begin{equation} \nu_g=1.15\cdot 10^{-3}\cdot 10.8\ \mathrm{ft^2/s}=12.4\cdot 10^{-3}\ \mathrm{ft^2/s}. \end{equation}

Now, since is the kinematic viscosity of glycerin is, by definition, equal to:

\begin{equation} \nu_g=\dfrac{\mu_g}{\rho_g}, \end{equation}

it follows that the viscosity of glycerin $\mu_g$ is given by:

\begin{equation} \mu_g=\nu_g\rho_g. \end{equation}

The density is related to the specific gravity by the following equation:

\begin{equation} S_g=\dfrac{\rho_g}{\rho_w}, \end{equation}

where $\rho_w=1940\ \mathrm{slug/ft^3}$ is the density of water (in FPS units). Therefore, $\rho_g=S_g\rho_w$ , which, since $S_g=1.26$ , can be plugged into equation (1) to obtain: