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 = 0.145 kg located at ?1 = 28.5° and a second mass m2 = 0.219 kg located at ?2 = 275°. Calculate the mass m3, and location (in degrees), ?3, which will balance the system and the ring will remain stationary.

Answer 1

m1 = 0.145kg. Weight = (0.145 x 9.81) = 1.422N.
m2 = 0.219kg., = 2.15N. weight.
Let's "rotate" the table 28.5 degrees anticlockwise, so m1 is at 0. That puts m2 at (275 - 28.5) = 246.5 degrees.
(246.5- 180) = 66.5 degrees "west of south".
South component of m2 = (cos 66.5) x 2.15, = 0.857N.
West component = (sin 66.5) x 2.15, = 1.971N.
Subtract South component from weight m1, = 0.565N., acting North.
Weight of M3 = sqrt. (1.971^2 + 0.565^2), = 2.05N., divided by g = mass of 0.21kg., or 210g.
Direction = arctan (0..565/1.971) = 16 degrees S of E.,
Now "rotate" the table back the 28.5 degrees clockwise, (28.5 + 16 + 90) = 134.5 degrees, is where to place the m3 of 210 g. (The "90" is because east is 90 deg. from North, and the 16 was from E).