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 = 0.161 kg located at 1 = 26.5° and a second mass m2 = 0.219 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.

Explanation / Answer

m1 = 0.161kg = (0.161 x 9.81) = 1.58N

m2 = 0.219kg = 2.15N

Let's "rotate" the table 26.5 degrees anticlockwise, so m1 is at 0. That puts m2 at (291 - 26.5) = 264.5 degrees.

(251.8 - 180) = 84.5 degrees "west of south".

South component of m2 = (cos 84.5 ) x 2.15 = 0.21N.

West component = (sin 84.5) x 2.15 = 2.14N.

Subtract South component from weight m1, = 1.37N acting North.

Weight of M3 = sqrt. (2.14^2 + 1.37^2), = 2.54N = 0.259 kg = 259 g

Direction = arctan (1.37/2.14) = 32.6 degrees S of E

Now rotate the table back the 26.2 degrees clockwise, (26.5 + 32.6 + 90) = 149.1 degrees, is where to place

the m3 of 259g

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