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Physics, 25.11.2020 02:00 shhells

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Physics, 22.06.2019 00:30, Solany6426
Part f - example: finding two forces (part i) two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. the block in (figure 1) has a mass m=10kg and is being pulled by a force f on a table with coefficient of static friction îľs=0.3. four forces act on it: the applied force f (directed î¸=30â above the horizontal). the force of gravity fg=mg (directly down, where g=9.8m/s2). the normal force n (directly up). the force of static friction fs (directly left, opposing any potential motion). if we want to find the size of the force necessary to just barely overcome static friction (in which case fs=îľsn), we use the condition that the sum of the forces in both directions must be 0. using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: fcosî¸â’îľsn=0 fsinî¸+nâ’mg=0 in order to find the magnitude of force f, we have to solve a system of two equations with both f and the normal force n unknown. use the methods we have learned to find an expression for f in terms of m, g, î¸, and îľs (no n).
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During the experiment if you could triple the breakaway magnetic force with all other quantities left unchanged, what is the new value for the critical velocity if it was v0 (initial velocity), initially? (b) now if you halved the radius with all other quantities left unchanged, what is the new critical velocity if it was v0 (initial velocity), initially? (c) if during the experiment, critical velocity quadrupled with all other quantities left unchanged, what is the new breakaway force if its magnitude was initially f0,?
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