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 = 0.153 kg located at 1 = 22.5° and a second mass m2 = 0.219 kg located at 2 = 295°. Calculate the mass m3, and location (in degrees), 3, which will balance the system and the ring will remain stationary.

Explanation / Answer

m1 = 0.153 kg. Weight = (0.153 x 9.81) = 1.5 N.
m2 = 0.219 kg., = 2.15 N. weight.
Let's "rotate" the table 22.5 degrees anticlockwise, so m1 is at 0. That puts m2 at (295 - 22.5) = 272.5 degrees.
(272.5 - 180) = 92.5 degrees "west of south".
South component of m2 = (cos 92.5) x 2.15 = -0.094 N.
West component = (sin 92.5) x 2.15 = 2.15 N.
Subtract South component from weight m1, = 1.594 N acting North.
Weight of M3 = sqrt. (2.15^2 + 1.594^2), = 2.68 N., divided by g = mass of 0.273 kg., or 273 g.
Direction = arctan (1.594/2.15) = 36.55 degrees S of E.,
Now "rotate" the table back the 22.5 degrees clockwise, (22.5 + 36.55 + 90) = 119.05 degrees, is where to place the m3 of 273 g. (The "90" is because east is 90 deg. from North, and the 36.55 was from E)

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