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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).

Which of the following bonds would be most polar? a. c-i b. c-br c. c-cl d. c-f e. c-o

Steelhead trout migrate up stream to spawn.occasionally they need to leap up small waterfalls to continue their journey. fortunately, steelhead are remarkable jumpers, capable of leaving the water at a speed of 8.0 m/s. a. what is the maximum height a steelhead trout can jump? b.leaving the water at 8.0 m/s, the trout lands on top of the water fall 1.8 m high. how long was it in the air?