You have a balance beam, the kind of scale that tips from one side to the other, depending on the weight on each side. On each side is a beaker, half-filled with water. The sides are in balance. Now, on the left side, you submerge a Ping-Pong ball suspended by a string. On the right side, you submerge a steel ball of the same volume as the Ping-Pong ball suspended from a crane.
You may ignore the mass of the strings.
Does the balance beam tip to the right, to the left, or does it remain unchanged?
The difference between forces of tension is that the string tension is an internal force in the beaker/water/ball/string system on the left, and the tension is an external force in the beaker/water/ball/(part of the string) system on the right.
Let’s write equation for the right beaker/water (without ball) system.
NR = Fa + Wbw , where NR – the normal force on the right system upward, Fa – the buoyancy force(Archimedes force) from the steel ball on the water (according the Newton’s third law) downward and Wbw – weight of the beaker/water downward.
Now let’s write equation for the left beaker/water/ball system.
NL= Wpb+Wbw, where NL – the normal force on the left system upward, Wpb – weight of the Ping-Pong ball downward and Wbw – weight of the beaker/water downward.
Compare the normal forces (We could compare this forces because the similar forces act on left and right sides of the balance beam.)
NR – NL = Fa – Wpb. Obviously Fa bigger than Wpb. Thereby, NR bigger NL.
The balance beam tip to the right.