Question

Vectors Equipment Force table 3 pulley clamps Masses and hangers Thread Procedure l. Assemble the force table by attaching the legs. The center post should be flush with the table top. 2. Take two threads (each about 60 cm long) and knot them together in the center, making four shorter threads. Pull one of the threads through the center post and tie it to one of the legs. This is the anchor string. 3. Attach mass hangers to the other three threads by winding the threads around the top of the hangers several times. 4. Hang the following masses on two of the pulleys and clamp the pulleys at the given angles: Mass A: 50 g at 0 Mass B: 100 g at 120 Adjust the pulleys downward until the strings are close to (but not touching) the top surface of the force table. 5. By trial and error, find the angle for the third pulley and the mass which must be suspended from it which will balance the forces exerted by the other two masses. (Hint: find the location of the third pulley first, then find the mass). When equilibrium is reached, the knot on the string will be directly over the hole in the center of the table. (If you pull the knot slightly to one side, it should return to the center.) Adjust the mass/position until equilibrium is reached, and record the mass and angle. 2 6. Repeat steps 4-5 for the following masses: 7S gramS Mass B: 100 g at 60° ISO 9

Answer 1

Analysis 1. Trial 1 m_a = 50 g, theta = 0 degree m_b = 100g, theta_b = 120 degree m_c = 75g, theta_c = 285 degree Trial 2 m_a = 50 g, theta = 30 degree m_b = 100g, theta_b = 60 degree m_c = 75g, theta_c = 232 degree a) using unit vector notation Trial 1-F_a = m_a times g (cos 0 degree i + sin 0 degree j^) F_b = m_b times g (cos 120 degree i^+ sin 120 j^) F_c = m_c times g (cos 285 degree i^+ sin 285 degree j^) For unknown F_c = ? F_a + F_b = -F_c hence F_c = -50 times g (i^) - 100 times g(0.814 i^+ 0.58 j^) but F_c = m_c times g m_c = -50 j^- 100(0.814 i^+ 0.58 J^) = -50 i^- 814 i^- 58 j^= -1314 i^- 58 j^ for trial 2, similarly