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 = 0.161 kg located at θ1 = 22.5° and a second mass m2 = 0.215 kg located at θ2 = 283°. Calculate the mass m3, and location (in degrees), θ3, which will balance the system and the ring will remain stationary.

Answer 1

m1 = 0.161kg. Weight = (.157 x 9.81) = 1.579N.
m2 = 0.215kg., = 2.109N. weight.
Let's "rotate" the table 22.5 degrees anticlockwise, so m1 is at 0. That puts m2 at (283 - 22.5) = 260.5 degrees.
(260.5 - 180) = 62.5 degrees "west of south".
South component of m2 = (cos 62.5) x 2.109, = 0.9738N.
West component = (sin 62.5) x 2.109, = 1.870N.
Subtract South component from weight m1, = 0.6052N., acting North.
Weight of M3 = sqrt. (1.870^2 + 0.6052^2), = 1.9056N., divided by g = mass of 0.194kg., or 194g.
Direction = arctan (0.6052/1.870) = 17.933 degrees S of E.,
Now "rotate" the table back the 22.5 degrees clockwise, (22.5 + 17.933 + 90) = 130.433 degrees, is where to place the m3 of 194g. (The "90" is because east is 90 deg. from North, and the 17.933 was from E).