If i exert a force $\vec{F_1}$ on a domino it would fall onto the second hitting it with a force $\vec{F_2}$, my question is : Would $\vec{||F_2||}$ be equal to $||\vec{F_1}|| + m.a$ or would $\vec{F_1}$ wither away during the fall? Or is my reasoning totally off?
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$\begingroup$ Not clear. What is $a$ in your question? Is the dot in $m.a$ a scalar product? What exactly is your reasoning here? ie Why do you think $\vec{||F_2||}$ would equal $||\vec{F_1}|| + m.a$? Why would $\vec{||F_1||}$ wither away during the fall? I think you have some major misconceptions about how forces work during and after collisions. $\endgroup$– sammy gerbilCommented Sep 17, 2018 at 13:09
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$\begingroup$ acceleration and no not a dot product. my reasoning was bad i guess, i thought about it again and probably the force 1 is defined with the initial acceleration due to it when the domino first moves. $\endgroup$– user184836Commented Sep 19, 2018 at 14:08
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1$\begingroup$ See math.udel.edu/~rossi/Math512/2005/Team3.pdf for an example of theory and experiments. $\endgroup$– sammy gerbilCommented Sep 19, 2018 at 20:18
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2 Answers
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Dominos typically rotate when they fall, so you should reason in terms of torque and take the angle of the domino on impact into account. So, no, F2 wont be equal to F1 +m.a.
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1$\begingroup$ I've seen the notation before, it means magnitude of vector. $\endgroup$ Commented Sep 14, 2018 at 7:49
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$\begingroup$ If you don't know what symbols in a question mean your should not try answering it until you ask the OP (in a comment) or find out on your own. $\endgroup$ Commented Sep 14, 2018 at 7:50
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$\begingroup$ cant comment I dont have enough reputation, tried google unsuccessfully too $\endgroup$ Commented Sep 14, 2018 at 7:50
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$\begingroup$ couldnt find it here for ex :en.wikipedia.org/wiki/List_of_common_physics_notations $\endgroup$ Commented Sep 14, 2018 at 7:52
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$\begingroup$ but anyhow, involving the magnitude makes it even worse, so I'll keep my answer $\endgroup$ Commented Sep 14, 2018 at 7:55
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If you look at a row of dominoes falling it is clear that the speed of the fall becomes constant. That means that although more and more energy has been released the energy in the falling region is roughly constant. Hence there is a fair bit of dissipation of that energy and the dissapearance of the initial "force" (actually the activation energy)
The exact calculation of forces and torques is not particularly illuminating.
answered Sep 14, 2018 at 7:36