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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).

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges zero? express your answer in terms of some or all of the variables s, q1, q2 and k =14ï€ïµ0. if your answer is difficult to enter, consider simplifying it, as it can be made relatively simple with some work.

Carbon is allowed to diffuse through a steel plate 15 mm thick. the concentrations of carbon at the two faces are 0.65 and 0.30 kg c/m^3 fe, which are maintained constant. if the preexponential and activation energy are 6.2 x 10-7 m2 /s and 80,000 j/mol, respectively, compute the temperature at which the diffusion flux is 1.43 x 10^-9 kg/m^2 -s.