F1=m*g = 0.151[26.2o]*9.8=1.480N[26.2o]
F2 = 0.215[281o]*9.8 = 2.107N[281o]
F3[Ao]
F = F1 + F2 + F3
F=1.480N[26.2o] + 2.107N[281] + F3[Ao]
X = 1.480*cos26.2+2.107*cos281+F3*cosA=0
X = 1.328 + 0.4020 + F3*cosA = 0
F3*cosA = -1.73
cosA = -1.73/F3.
Y = 1.480*sin26.2+2.107*sin281+F3*sinA=0
Y = 0.6534 - 2.068 + F3*sinA = 0
F3*sinA = 1.415
sinA = 1.415/F3
sinA/cosA = (1.415/F3)/(-1.73/F3) =
1.415/-1.73 = -0.81792 = TanA
A = -39.3o = 320.7o, CCW.
F3*sinA = 1.415
F3 = 1.415/sinA = 1.415/sin320.7=-2.23 N[320.7o]
m3*g = -2.23
m3=-2.23/g = -2.23/-9.8=0.228 kg[320.7o]
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 held in place with a pin. 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 of each string. There is a mass of m1 = 0.151 kg located at è1 = 26.2° and a second mass of m2 = 0.215 kg located at è2 = 281°. Calculate the mass m3, and location (in degrees), è3, which will balance the system so the pin can be removed and the ring will remain stationary. The acceleration due to gravity is g = 9.81 m/s2.
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