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In the below problem the weight of the rod acts at it's centre (the centre of mass). But how do I know from the buoyant force acts? How do I proceed with the rest of the problem?

The Problem is shown in the image URL

Here's my attempt to answer it:

Since I can assume that the rod is hinged about the point where it is bound with the string... The net external torques about the hinge point should be equal to zero.

However, for me to do that I need to know the distance from the hinge point to where the buoyant force is acting on which I am stuck.

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  • $\begingroup$ The weight is the cumulative force of the gravitation acting on all parts of the body. So it doesn't act on the center of mass, but it is equivalent to a point force acting on the center of mass. The Buoyant Force is the cumulative effect of the pressures acting on all external surfaces of the body. If you want to assign it an equivalent point force, you can take it through the center of mass also. $\endgroup$ – Chet Miller Sep 29 at 12:19
  • $\begingroup$ I have edited the answer kind sir ... $\endgroup$ – user200080 Sep 29 at 12:54
  • $\begingroup$ As @R.W. Bird points out, they expect you to assume that the buoyant force is directed vertically, and at the center of the submerged section. I think I would more comfortable with this assumption if I integrated the pressure distribution on the submerged section to confirm this. $\endgroup$ – Chet Miller Sep 29 at 21:12
  • $\begingroup$ The buoyant force must be vertical because before the rod is inserted, that same force must support the weight of the water that will be displaced. $\endgroup$ – R.W. Bird Sep 30 at 17:03
  • $\begingroup$ It does occur to me that the fact that the cut through the rod at the surface is not parallel to the lower end of the rod may disturb the assumption of uniformity for the buoyant force. $\endgroup$ – R.W. Bird Sep 30 at 17:13
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Keeping in mind that the buoyant force depends on the difference between the water pressure below and above the rod, then the buoyant force acts uniformly along the rod and the effective force can be taken at the middle of the submerged section. With the force equation: buoyant force + tension = weight, and and a torque equation (choose an axis), you can solve for x/L, the fraction above water. I got x = L/2 (which works when placed back in the two equations).

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  • $\begingroup$ That's the correct answer but you only need the torque equation for that .. Regardless L/2 is perfectly correct $\endgroup$ – user200080 Sep 30 at 15:28

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