To better understand the concept of static equilibrium a laboratory procedure as

Question

To better understand the concept of static equilibrium a laboratory procedure asks the student to make a calculation before performing the experiment. The apparatus consists of a round level table in the center of which is a massless ring. There are three strings tied to different spots on the ring. Each string passes parallel over the table and is individually strung over a frictionless pulley (there are three pulleys) where a mass is hung. The table is degree marked to indicate the position or angle of each string. There is a mass m1 = 157 g located at ?1 = 22.5

Explanation / Answer

m1 = 157g = 0.157kg. Weight = (.157 x 9.81) = 1.54N.
m2 = 0.215kg., = 2.109N. weight.
Let's "rotate" the table 22.5 degrees anticlockwise, so m1 is at 0. That puts m2 at (275 - 22.5) = 252.5 degrees.
(252.5 - 180) = 72.5 degrees "west of south".
South component of m2 = (cos 72.5) x 2.109, = 0.6342N.
West component = (sin 72.5) x 2.109, = 2.0114N.
Subtract South component from weight m1, = 0.9058N., acting North.
Weight of M3 = sqrt. (2.0114^2 + 0.9058^2), = 4.866N., divided by g = mass of 0.229kg., or 229g.
Direction = arctan (0.9058/2.0114) = 24.244 degrees S of E.,
Now "rotate" the table back the 26.2 degrees clockwise, (26.2 + 24.244 + 90) =140.444 degrees, is where to place the m3 of 229g. (The "90" is because east is 90 deg. from North, and the 24.244 was from E).

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