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We assume that the object's density is the same everywhere, and that indeed has a perfectly rectangular shape. We also assume that its volume is continuous and dont work with particles here.

To calculate the torque of the object, you must multiply the distance to the fixed point A times the force exerted by it. The easiest way to do it is to multiply the distance between A and the center of gravity, and the object's weight $F_1$. The torque thus equals $F_1 2m$ (m stands here for meter, not mass). I also wanted to see if there is an other solution to this problem that doesnt involve the COG, but instead treats the object as a whole, and takes its area into consideration. The second approach, explained further below, should give the same solution as the first one, as both talk about the same object with the same placement.

The second solution involves dividing the object into infinitely small columns of mass, calculating the torque of each column and finally calculating their sum. Each column is equally wide and has the same height, thus they must have the same mass -> gravitational force. Lets call the column's gravitational force $F_2$ (each column will have the same gravitational force). The torque of each column will be calculated by multiplying its distance to the point A times its weight $F_2$. We can thus create a function where the torque of each column will be its weight $F_2 x$ (the distance). To sum all the torques, you must integrate the function over [1.5, 2.5] giving:

$\frac {(2.5m × 2.5m × F2 - 1.5m × 1.5m × F2)}{2} = 2m^2 × F2$. How do I proceed? There is a $m^2$ and I have thus no idea how to continue.

I also tried just removing the m, but then I get that the sum of all the torques equals $2F_2$, which would mean that for both solutions to be equal, the weight of one column has to equal the weight of the whole object which seems very unreasonable.

  • $\begingroup$ Please use the MathJax syntax to make your mathematical formulas easier to read. Please also explain, what the symbol $m$ in your post means. Is it a length unit or a mass? $\endgroup$ – flaudemus Mar 5 at 10:18
  • $\begingroup$ m stands here for meter, not mass $\endgroup$ – DaddyMike Mar 5 at 10:27

Let the distance of a column from $A$ be $x$ and there are $n=\infty$ colums, the sum function becomes: $$\sum^{\infty}_{n=1} \sum^{2.5}_{x=1.5} F_2x$$

The integral becomes: $$F_2 \int^{2.5}_{1.5}x dx$$

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Mar 6 at 18:04

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