Lesson 2.2 Motion of Connected Masses

# Lesson 2.2 Motion of Connected Masses - Connected Masses In...

This preview shows pages 1–7. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 Since the masses are connected, each mass has the same acceleration, namely 3 ms -2 . A total force F net = 18 N pulls a total mass m = 1 + 2 + 3 = 6 kg net F a m = 18 N 6 kg = = 3 ms -2 . 2
m 1 T 1 1.0 kg m 2 = 2 kg m 2 = 2 kg T 2 T 1 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 m 2 T – 0.1a = 0.98 ..(i) T – 0.98 = 0.1a If a is the acceleration of the system, then The tension T on the string acts vertically up. T m 1 g 4
m 2 = 2 kg T = 0.985 N and a = 0.05 ms -2 Solving these equations give: T + 0.101a = 0.99 T – 0.1 a = 0.98 Now we have two equations: 0.101a + T = 0.99 …(ii) The tension T acts vertically up m 1 m 2 m 1 g T m 2 g T T – 0.1a = 0.98 ..(i) 5

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Consider a mass m resting on a plane surface. m 2 = 2 kg Object on an Inclined Plane m mg m θ mg A line drawn at right angles to the plane makes angle θ with mg θ The vector component of mg at right angles to the plane is mg cos θ mg cos θ mg sin θ The component of mg down the plane is mg sin θ The other part pulls it down the plane.
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 17

Lesson 2.2 Motion of Connected Masses - Connected Masses In...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online