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 = 0.157 kg located at θ1 = 26.5° and a second mass m2 = 0.215 kg located at θ2 = 275°. Calculate the mass m3, and location (in degrees), θ3, which will balance the system and the ring will remain stationary.

Answer 1

given ,

m1 = 0.157 kg

m2 = 0.215 kg

theta 1 = 26.5 degrees

theta 2 = 275 degrees

w1 = mass1*gravity

= 0.157*9.8

=1.539 N

w2 = mass2*gravity

= 0.215*9.8

= 2.107 N

m2 angle= (theta 2 - theta 1)

= (275 - 26.5)

= 248.5 degrees

South west = (248.5 - 180)

= 68.5 degrees

West component = (sin 68.5)*w2

=  (sin 68.5) x 2.107

= 1.96N

South component of m2 = (cos 68.5) * w2

= (cos 68.5) * 2.107

= 0.772 N.

North = W1 - South component

= 1.539 - 0.772

= 0.767N.,

W3 = sqrt. (1.96^2 + 0.767^2),

= 2.105N

w3 / g = 2.105 / 9.8

= 215 g
theta = tan^-1 (0.767 / 1.96)

= 21.4 degrees (south of east)
theta 1 in clockwise, (26.5 + 21.4 + 90) = 137.9 degrees

m3 = 215g.

90 degrees from North

21.4 degrees from East