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 = 0.157 kg located at ?1 = 24.5? and a second mass m2 = 0.211 kg located at ?2 = 291?. Calculate the mass m3, and location (in degrees), ?3, which will balance the system and the ring will remain stationary.

To better understand the concept of static equilibrium a laboratory procedure asks the student to make a calculation before performing the experiment. The apparture consists of a round level table in the center of which is a massless ring. These 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 in hung. The table is degree marked to indicate the position or angle of eash string. There is a mas m1 = 0.157 kg located at theta1 = 24.5 degree and a second mass m2 = 0.211 kg located at theta2 = 291 degree . Calculate the mass m3, and location (in degrees), theta3, which will balance the system and the ring will remain stationary. m3 = kg theta = degree

Explanation / Answer

here

m1 = 0.157kg. weight = (.157 x 9.8) = 1.543N

m2 = 0.211kg. weight = 2.06 N

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

South component of m2 = (cos 86.5) * 2.06 = 0.1257N
West component = (sin 86.5 ) * 2.06 = 2.0561N

Subtract South component from weight m1, = 1.543 - 0.1257 = 1.41 N acting North

Weight of M3 = sqrt. (2.056^2 + 1.41^2), = 2.49N divided by g = mass of 0.254kg or 254g

Direction = arctan (1.41 / 2.056) = 34.44 degrees S of E

Now "rotate" the table back the 26.2 degrees clockwise, (24.5 + 34.44 + 90) = 148.94 degrees, is where to place the m3 of 254g. (The "90" is because east is 90 deg. from North, and the 34.44 was from E)

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