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 = 28.5° and a second mass m2 = 0.219 kg located at ?2 = 279°. Calculate the mass m3, and location (in degrees), ?3, which will balance the system and the ring will remain stationary.

sapling leaming 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 asses parallel over the table and is individually strung over a rictionless 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 m = 0.149 kg located at -28.5 and a second mass m2 = 0.219 kg located at 2-279" Calculate the mass m3, and location (in degrees), , which will balance the system and the ring will remain stationary Number m = 0.219 Number

Explanation / Answer

m1 = 0.149kg. Weight = (.149 x 9.81) =1.46N.

m2 = 0.219kg*9.81., = 2.109N. weight.

Let's "rotate" the table 28.5 degrees anticlockwise, so m1 is at 0. That puts m2 at (279 - 28.5) = 250.5 degrees.

(250.5 - 180) = 70.5 degrees "west of south".

South component of m2 = (cos 70.5) x 2.109, = 0.70399N.

West component = (sin 70.5) x 2.109, = 1.9880N.

Subtract South component from weight m1, = 0.70399N., acting North.

Weight of M3 = sqrt. (1.9980^2 + 0.70399^2), = 2.118N., divided by g = mass of 0.229kg.,

Direction = arctan (0.70399/1.9980) = 19.37 degrees S of E.,

Now "rotate" the table back the 28.5 degrees clockwise, (28.5 + 19.37+ 90) = 137.879 degrees, is where to place

the m3 of .229kg. (The "90" is because east is 90 deg. from North, and the 19.37 was from E).

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