Atwood Machines and Multiple Body Systems
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.
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
Find acceleration and tension 3.0 kg 5.0 kg
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 = 0.5 b) a=1.96 m/s2 c) 39.25 N
Find acceleration and tension if μ=.05 11.0 kg Find acceleration and tension if μ=.05 30.0o 6.0 kg +a
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.
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 0.200. Determine the acceleration and the contact force.
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?
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)