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 = 0.149 kg located at 1 = 22.5° and a second mass m2 = 0.215 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.

Explanation / Answer

Given,

mass m1 = 0.149kg.

Weight = (0.149 x 9.81)

= 1.46 N.

m2 = 0.215kg

= 2.109N. weight.
Let's "rotate" the table 22.5 degrees anticlockwise

m1 is at 0. That puts m2 at (275 - 22.5) = 252.5 degrees.

(252.5 - 180) = 72.5 degrees "west of south".

South component of m2 is

=(cos 72.5) x 2.109,

= 0.6341N.

West component = (sin 72.5) x 2.109

= 2.0113N.

Subtract South component from weight m1, = 0.8259 N., acting North.

Weight of M3 = sqrt. (2.0113^2 + 0.8259^2)

= 2.174 N., divided by g = mass of 0.222kg., or 222g.

Direction = arctan (0.8259/2.0113)

= 22.324 degrees S of E

Now "rotate" the table back the 22.5 degrees clockwise, (22.5 + 22.324 + 90) = 134.824 degrees

is where to place the m3 of 222 g.

(The "90" is because east is 90 deg. from North, and the 22.324 was from E).

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