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In the image attached, what would be the force that would balance out the gravitational force $mg$?

The block is also not accelerating downwards so there should be some force acting on the block to balance out mg. As the tension is acting perpendicular to mg so there is no component of tension too.

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    $\begingroup$ There is no such thing as a perfectly rigid rod. In reality, that rod will bend slightly, giving rise to a vertical tension. Related: Rotating a stone with a string. $\endgroup$
    – lemon
    Commented May 27, 2016 at 9:39
  • $\begingroup$ It may help to know that the rule "tension acts along the line" only applies to strings, light ropes and other physics objects that bend very easily. The connecting bar here is described as a "rod". $\endgroup$ Commented May 27, 2016 at 16:12
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    $\begingroup$ Hmm ... I have to disagree with the homework-like votes on this one. The question isn't related to the solution to the exercise but to understanding the forces perpendicular to the ones the question asks about. $\endgroup$ Commented May 27, 2016 at 16:56
  • $\begingroup$ The question does not mention gravity. I am guessing that you added the arrows to the diagram. I think you are expected to assume this experiment takes place in a gravity-free environment. $\endgroup$ Commented Jun 1, 2016 at 12:29
  • $\begingroup$ Hi Osheen Sachdev. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$
    – Qmechanic
    Commented Jun 1, 2016 at 14:19

2 Answers 2

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The rod exerts an upward shear force on mass on the left. This goes along with @lemon's comment regarding bending, although the shear force is not a tension. There is also a bending moment applied by the rod to the mass, with axial tension within the top half of the rod, and axial compression within the bottom half.

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When rotating, the force needed to keep the two masses in circular motion is the centripetal force, (in this case, the tension). Now, you get it.

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  • $\begingroup$ This doesn't answer the question. The question asks about forces in the "y" direction, not forces in the "x" direction. $\endgroup$
    – Ian
    Commented May 27, 2016 at 18:36

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