Question 6.3.6: A solid steel ball of radius r = 50.0 mm and specific weight...

Goal Find the tension in the wire that is holding up the ball.

Given Properties of the steel ball and the liquid it is suspended in.

Assume The specific weights are uniform and the ball is at rest (in equilibrium).

Draw The hydrostatic forces acting in the horizontal direction sum to zero, so we have not included them on the free-body diagram of the steel ball (Figure 2).

Formulate Equations and Solve We apply equilibrium in the vertical direction:

T =W – F_{oil} – F_{Hg} (1)

where

W=\gamma _{ball} V_{ball}=\gamma _{s}\frac{4}{3 } \pi r^{3}=\frac{4}{3} \left(76.8  kN m^{3} \right)\pi \left(0.0500  m\right) ^{3} =40.21  N
F_{oil}=V_{ball  submerged} \left(\gamma _{oil}\right) =\frac{1}{2}V_{ball} \left(\gamma _{oil}\right)
=\frac{1}{2} \left\lgroup\frac{4}{3} \pi r^{3}\right\rgroup\gamma _{oil}= \frac{1}{2}\frac{4}{3}\pi \left(0.0500  m\right) ^{3} \left(7.80  kN/ m^{3}\right) =2.042  N
\gamma _{Hg}=V_{ball  submerged}\left(\gamma _{Hg}\right) =\frac{1}{2}V_{ball}\left(\gamma _{Hg}\right)
=\frac{1}{2}\left\lgroup\frac{4}{3}\pi r^{3}\right\rgroup\gamma _{Hg}= \frac{1}{2}\frac{4}{3}\pi\left(0.0500  m\right)^{3} \left(133  kN/m ^{3} \right) = 34.82  N

Find T using (1) and the calculated values of F_{oil} and F_{Hg} :

T =W – F_{oil} – F_{Hg} =40.21 N – 2.042 N – 34.82 N = 3.35 N

If T were negative, it would mean that the wire is in compression. A thin wire would not support a compression force, so it would buckle and the steel ball would rise. A downward force would therefore be needed to sink the steel ball halfway into each liquid.

Check Other than redoing the calculations, there is no good check for this problem.