Lesson_2.2_Printable

# Lesson_2.2_Printable - Connected Masses In order to do...

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Connected Masses In order to do problems using connected masses, 1. We isolate each object, 2. Identify the forces acting on each. 3. Apply the laws of motion to form equations in each case. 4. Solve the equations for the unknown. Three blocks are pulled along a frictionless horizontal surface by applying a force of 18 N as shown below. (a) What is the acceleration of the system? (b) What are the tension forces in the two pieces of strings connecting m 1 , m 2 and m 3 ? m 3 m 2 m 1 T 1 T 1 T 2 T 2 3.0 kg 2.0 kg 1.0 kg F = 18.0 N 1

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m 3 m 2 m 1 T 1 T 1 T 2 T 2 3.0 kg 2.0 kg 1.0 kg F = 18.0 N The tension on the string connecting masses m 1 and m 2 is T 1 ulls m the right and m the left. The tension on the string connecting masses m 2 and m 3 is T 2 Notice here that T 1 pulls m 1 to the right and m 2 to the left. Since the masses are connected, each mass has the same acceleration, namely 3 ms -2 . F net = ma A total force F net = 18 N pulls a total mass m = 1 + 2 + 3 = 6 kg T 2 pulls m 2 to the right and m 3 to the left. net F a m = 18 N 6 kg = = 3 ms -2 . 2
m 1 T 1 1.0 kg To find the tension T 1 , we isolate the mass m 1 m 2 = 2 kg m 2 = 2 kg T 2 T 1 Applying F net = ma on m 1 gives:. The tension T 1 produces an acceleration 3 ms -2 on m 1 = 1 kg. The net force acting on m 1 is T 1 . T 2 = T 1 + 6.0 N = 3.0 N + 6.0 N = 9.0 N T 2 – T 1 = 2.0 kg × 3.0 ms -2 = 6.0 N Applying F net = ma on m 2 gives: The forces on m 2 are T 2 to the right and T 1 to the left To find T 2 , we isolate m 2 T 1 = 1.0 kg × 3.0 ms -2 = 3.0 N 3

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Since the same string supports the two masses and there is no friction on the pulley, the tension on the right is the same as the tension on the left = T m 2 = 2 kg Two masses m 1 = 100 g and m 2 = 101 g are connected by a string that passes over a frictionless pulley. Find the acceleration of the masses and the tension on the string m 1 The weight of m 1 = m 1 g = 0. 1 kg × 9.8 s 0.98 N acts vertically down. We now isolate the mass m 1 = 100 g = 0.1 kg. T m 2 F net = m 1 a T – 0.1a = 0.98 ..(i) T – 0.98 = 0.1a If a is the acceleration of the system, then The net force on m 1 is T – m 1 g = T – 0.98 N Since m 2 > m 1 , we will assume that m 1 moves up and m 2 moves down. The tension T on the string acts vertically up. ms -2 = 0.98 N acts vertically down. m 1 g 4
To obtain another equation we isolate m 2 m 2 = 2 kg m 2 has the same acceleration a The net force on m 2 is m 2 g – T = 0.99 – T The tension T acts vertically up The weight of m 2 = m 2 g = 0.101 kg × 9.8 N = 0.99 N acts vertically down

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## This note was uploaded on 05/01/2011 for the course PHY 2048 taught by Professor George during the Fall '10 term at Edison State College.

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Lesson_2.2_Printable - Connected Masses In order to do...

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