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 = 0.153 kg located at θ1 = 28.5° and a second mass m2 = 0.227 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

T₁ = magnitude of tension in the string connected with hanging mass m₁ = m₁ g = 0.153 x 9.8 = 1.5 N

T₂ = magnitude of tension in the string connected with hanging mass m₂ = m₂ g = 0.227 x 9.8 = 2.2 N

Using equilibrium of force

+ + = 0

((T₁ Cos ) i + (T₁ Sin) j) + ((T₂ Cos ) i + (T₂ Sin) j) + = 0

((1.5 Cos28.5 ) i + (1.5 Sin28.5) j) + ((2.2 Cos275 ) i + (2.2 Sin275) j) + = 0

+ (1.51 i - 1.48 j) = 0

= - 1.51 i + 1.48 j

magnitude : sqrt((- 1.51)² + (1.48)²) = 2.11

since magnitude of tension T₃ balances mass m₃ , hence

T₃ = m₃ g

2.11 = m₃ (9.8)

m₃ = 0.215 kg

direction is given as

= 180 - tan^-1(1.48/(1.51)) = 135.6 deg