Question

A light quarter cylinder of radius $R$ and length $L$ is hinged at smooth axis $A$. Cylinder is supported by a horizontal rod. Liquid of density $\rho$ is filled between wall and cylinder. If tension in rod is $\frac{\rho g R^2 L}{\alpha}$. Find value of $\alpha$. (rod can apply force in horizontal direction only)

Answer 1

3

Answer 2

Solution:

We note that for any curved surface the horizontal component of the hydrostatic force equals the force on its vertical projection. Here the curved surface (the quarter‐cylinder) projects onto a vertical rectangular area of width R and the same length L. Thus, the net horizontal force is

$$F = \rho g h_c(A_p) = \rho g\left(\frac{R}{2}\right)(R\,L)=\frac{\rho g R^2L}{2},$$

since the centroid of a rectangle of height R (with the top at the free surface) lies at $R/2$ below the free surface.

However, the moment of this force about the hinge (A) must be computed about the center of pressure of the vertical surface. For a vertical plane with its top at the free surface and bottom at depth R, the center of pressure is at

$$y_{cp} = \frac{R}{2}+\frac{I}{y_cA} = \frac{R}{2} + \frac{(R^3/12)}{(R/2)(R)} = \frac{R}{2} + \frac{R}{6} = \frac{2R}{3}.$$

Thus, the moment about A due to the hydrostatic force is

$$M_F = F\;y_{cp} = \frac{\rho g R^2L}{2}\cdot \frac{2R}{3} = \frac{\rho g R^3L}{3}.$$

The rod provides a horizontal force $T$ whose line of action (assumed to act at the free end of the quarter‐cylinder) has a lever arm of R with respect to the hinge. For equilibrium about A we require

$$T\,R = M_F = \frac{\rho g R^3L}{3},$$ $$\implies T = \frac{\rho g R^2L}{3}.$$

Comparing with the given expression

$$T = \frac{\rho g R^2L}{\alpha},$$

we identify

$$\alpha = 3.$$