Question 25 of 33 Sapling Learning To better understand the concept of static eq

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Question 25 of 33 Sapling Learning To better understand the concept of static equilibrium a laboratory before performing the experiment. The apparatus consists of a round level table in the center of which is a massless ring. There are three s ure asks the student to make a calculation tied to different spots on the ring. Each string asses parallel over the table and is individually strung over a ctionless 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 = 0.153 kg located at .-28.5. and a second mass m2 = 0.215 kg located at ½ = 295.. Calculate the mass ms and location (in degrees), , which will balance the system and the ring will remain stationary. Number kg Number O Previous Gve Up & View Solution Check Answer 0 Next Ext 2- Hint

Explanation / Answer

here

m1 = 0.153 kg. weight = (.153 x 9.8) = 1.5 N

m2 = 0.215 kg. weight = 2.11 N

Let's rotate the table 28.5 degrees anticlockwise, so m1 is at 0. That puts m2 at (295 - 28.5) = 266.5 degrees

(266.5 - 180) = 86.5 degrees west of south

South component of m2 = (cos 86.5) * 2.11 = 0.1288 N

West component = (sin 86.5 ) * 2.11 = 2.106 N

Subtract South component from weight m1, = 1.5 - 0.1288 = 1.37 N acting North

Weight of M3 = sqrt. (2.106^2 + 1.37^2), = 2.51 N divided by g = mass of 0.256 kg or 256 g

Direction = arctan (1.37 / 2.106) = 33.04 degrees S of E

Now "rotate" the table back the 26.2 degrees clockwise, (28.5 + 33.04 + 90) = 151.54 degrees, is where to place the m3 of 256 g. (The "90" is because east is 90 deg. from North, and the 33.04 was from E)

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