A laboratory on the concept of static equilibrium uses an apparatus called a force table. The force table consists of a round, level platform marked in degree increments about its circumference. In the center of the table sits a massless ring, to which three strings are tied. Each string is strung over a separate pulley clamped to the edge of the platform, and then tied to a freely-hanging mass, so that each string is under tension and the ring is suspended parallel to the table surface. A mass of ny = 0.157 kg is located at 6; = 24.5",and a second mass of m, = 0.211 kg is located at 8, = 275. Calculate the mass my and the angular position 85 (in degrees) that will balance the system and hold the ring stationary over the center of the platform. & TOOLS Y YA crate hangs from a rope that is attached to a metal ring. The metal ring is suspended by a second rope that is attached overhead at two points, as shown. What is the angle O if the tension in rope 1 is 1.19 times the tension in rope 2? # TOOLS x10\" Problem 2: Angle Calculation in Rope Tension Given: The tension in rope 1 (17) is 1.19 times the tension in rope 2 (153). To find: The angle 6. Solution: Given: T, =1.1975 The equilibrium conditions for the system require balancing the forces in both horizontal and vertical directions. Let's denote: e T as the tension in each of the two segments of the rope that are attached overhead. + T as the tension in the single rope that hangs vertically from the ring. Since T7 is directly downward and T5 tensions are symmetric, the horizontal components of the tensions in the upper ropes will balance each other. Therefore, we only need to consider the vertical components of these tensions. Vertical Force Balance: The vertical component of the tensions in the upper ropes must equal the tension in the lower rope: 272 cos 0 = Ti Given that Ti = 1.1972: 272 cos 0 = 1.1912 Dividing through by T2: 2 cos 0 = 1.19 Solving for cos 0: cos 0 = 1.19 2 = 0.595 Thus: 0 = arccos(0.595) Calculating the angle: 0 ~ arccos(0.595) ~ 53.4 Answer: 53.40