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 differnent spots on the ring.each string passes parallel over the table and is individually strung over a frictionless pulley (there are 3 pulleys) where mass is hung. The table is degree markes to indicate the position or the angle of each string. There is a mass m1= 0.157 kg located at theta1= 24.5 and a second mass m2=0.219kg located at theta2= 291 degrees. Calcualte the mass m3 and location in degrees theta 3, which will balance the system and the ring will remain stationary

Explanation / Answer

m1 = 0.157kg. Weight = (.157 x 9.81) = 1.54 N.
m2 = 0.219 kg = 2.148 N. weight.

Let's "rotate" the table 24.5 degrees so m1 is at 0. That puts m2 at (291 - 24.5) = 266.5 degrees.
(266.5 - 180) = 86.5 degrees "south of west".

South component of m2 = (cos 86.5) x 2.148, = 0.1311 N.
West component = (sin 86.5) x 2.148, = 2.143 N.
Subtract South component from weight m1, = 1.54-0.1311= 1.4089 N., acting North.
Weight of M3 = sqrt. (2.143^2 + 1.4089^2), = 2.564 N., divided by g = mass of 0.261 kg., or 261 g.
Direction = arctan (1.4089/2.143) = 33.32 degrees

Now "rotate" the table back the 24.5 degrees, (33.32 + 24.5 + 90) = 147.82 degrees, is where to place the m3 of 261 g. (The "90" is because east is 90 deg. from North, and the 33.32 was from E).

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