The empirical formula of (NaSO)​

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).

The rigid beam is supported by the three suspender bars. bars ab and ef are made of aluminum and bar cd is made of steel. if each bar has a cross-sectional area of 450 mm2, determine the maximum value of p if the allowable stress is (σallow)st = 200 mpa for the steel and ( σallow)al = 150 mpa for the aluminum. est = 200 gpa and eal = 70 gpa.

An object of dimensions 50 cm x 40 cm x 0.20 cm has a mass 60g. find its density in g/cm3 and kg/ m3