0
$\begingroup$

Please explain Michael Seifert's answer here

  • I think $x,y$ are distances but what is $d$?

  • How did he get this formula?

x cos ⁡ θ + d sin ⁡ θ
  • Does this equation implies Mgxcosθ= mgycosθ?
τ = − M g ( x cos ⁡ θ + d sin ⁡ θ ) + m g ( y cos ⁡ θ − d sin ⁡ θ ) = − ( M + m ) g d sin ⁡ θ
  • What does negative torque and positive θ displacement mean?

Any help would be wholeheartedly appreciated.

$\endgroup$
3
  • 1
    $\begingroup$ $x,y$ are horizontal distances. $d$ is the vertical height of the rod. Look at the diagram, he drew it in. $\endgroup$ Commented Dec 12, 2023 at 4:23
  • $\begingroup$ @Ghoster Fixed. $\endgroup$
    – gandalf61
    Commented Dec 12, 2023 at 7:47
  • $\begingroup$ I also mention in that answer that "positive rotations are clockwise." Similarly, counterclockwise torques and angular displacements are treated as negative. $\endgroup$ Commented Dec 12, 2023 at 12:47

1 Answer 1

0
$\begingroup$

$d$ is the vertical thickness of the beam. The answer is pointing out that although an ideal one-dimensional beam (with $d=0$) will be in neutral equilibrium, a real beam (with $d>0$) will actually be in stable equilibrium. This is because if a beam with non-zero thickness is displaced by a small angle $\theta$ from the horizontal, the resulting torque $\tau = - (M+m)gd \sin \theta$ will be in the opposite direction to $\theta$ and so acts as a restoring torque.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.