1. Atwood Machines and Multiple Body Systems

2. Atwood Machine FT m1g m2g m1 m2 2) a is the same for both masses.
The reason for using an Atwood machine is to provide a simple way to produce a small acceleration compared to the acceleration due to gravity m1 m2
m1g FT
m2g
We assume the rope & pulley have negligible mass, the rope doesn’t stretch, and the pulley is frictionless
KEY FACTS:
1) FT is the same throughout the rope.
2) a is the same for both masses.

3. Steps Draw free body diagram for each object
Write a Newton’s 2nd Law equation for each object, Fnet will be a different direction for each object
Solve for FT for each object
Set equations equal to each other and solve for a

4. Find acceleration and tension
3.0 kg
5.0 kg

5. Example a) ms = 0.5 b) a=1.96 m/s2 c) 39.25 N
Given m1 = 10 kg and m2 = 5 kg:
a) What value of ms would stop the block from sliding?
b) If the box is sliding and mk = 0.2, what is the acceleration?
c) What is the tension of the rope?
a) ms = b) a=1.96 m/s2 c) N

6. Find acceleration and tension if μ=.05
11.0 kg
Find acceleration and tension if μ=.05
30.0o
6.0 kg +a

7. 8 kg
5 kg
6 kg T1
frictionless floor T2
38 N
When there are more than 2 bodies or the bodies are connected, it is best to find the acceleration for the whole system first, then look at each object separately starting with the one the farthest away from the force.

8. A 5. 0-kg and a 10. 0-kg box are touching each other. A 45
A 5.0-kg and a 10.0-kg box are touching each other. A 45.0-N horizontal force is applied to the 5.0-kg box in order to accelerate both boxes across the floor. The coefficient of kinetic friction is Determine the acceleration and the contact force.

9. Box / Tension Problem What is the tension in each rope? T2 T1 8 kg
frictionless floor T2
38 N
What is the tension in each rope?

10. Problem-Solving Hints Newton’s Laws
Draw a free body diagram
Break any forces at angles into their components
No acceleration - set forces on one side = forces on the other side
Acceleration – Write Fnet = bigger force(s) – smaller force(s)