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 m_1 = 0.145 kg located at theta_1 = 24.5^degree and a second mass m_2 = 0.211 kg located at theta = 287^degree. Calculate the mass m_3, and location (in degrees), theta_3, which will balance the system and the ring will remain stationary.

Explanation / Answer

m1 = 0.145 kg. Weight = (0.145 x 9.81) = 1.42 N.
m2 = 0.211 kg., = 2.07 N. weight.
Let's "rotate" the table 24.5 degrees anticlockwise, so m1 is at 0. That puts m2 at (287 - 24.5) = 262.5 degrees.
(262.5 - 180) = 82.5 degrees "west of south".
South component of m2 = (cos 82.5) x 2.07, = 0.2702 N.
West component = (sin 82.5) x 2.07 = 2.052 N.
Subtract South component from weight m1, = 1.1498 N acting North.
Weight of M3 = sqrt. (2.052^2 + 1.1498^2), = 2.352 N., divided by g = mass of 0.2398 kg., or 239.8 g.
Direction = arctan (1.1498/2.052) = 29.26 degrees S of E.,
Now "rotate" the table back the 24.5 degrees clockwise, (24.5 + 29.26 + 90) = 143.76 degrees, is where to place the m3 of 239.8 g. (The "90" is because east is 90 deg. from North, and the 29.26 was from E).

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