Newton's third law and free body diagrams

[kathyt.25]

Homework Statement

"Three blocks (m1, m2 and m3) are in contact with each other on a frictionless, horizontal surface. A horizontal force is applied to m1.
http://i4.photobucket.com/albums/y111/kathy_felldown/sb-pic0556.png
What is the net force on block 1? "

When drawing free body diagrams for this system, I know that you are only supposed to include forces acting upon THAT object, but I wasn't sure if I always needed to take Newton's third law into account. For example, would the FBD for object m1 have these forces:
- weight pointing down
- normal pointing up
- applied force pointing right
- force exerted by m2 ON m1... but I have no idea how to calculate this value!

What confuses me about Newton's third law is that if there is always an equal and opposite force, wouldn't that mean that the two objects that are in contact, wouldn't move at all even if a force is applied in one direction? Because isn't the force that's applied on m2 by m1 equal and opposite to the applied force, since the applied force is what is "forcing" m1 to move m2? Or is the force exerted on m2 by m1 NOT EQUIVALENT to the applied force?

Homework Equations

Fnet = ma

The Attempt at a Solution

I found acceleration by calculating the Fnet = m(total)*a
a = 17.9 / (1.61 + 3.39 + 4.16) = 1.95m/s/s

Just for curiosity, I tried the Fnet = (1.61kg)(1.95m/s/s) = 3.14N and it was correct.
I have no idea why the applied force, 17.9N is not considered in this... OR the force of m2 on m1 (according to Newton's 3rd law)

 

[davieddy]

You seem to have identified all the forces exerted ON m1.
These are what you need to add when applying Newton's 2nd
Law to m1.
The force exerted by m1 on m2 is NOT a force exerted on m1!

 

[nlightner]

is not included in the net force calculation.


I would like to address your confusion about Newton's third law and free body diagrams. First, let's review Newton's third law: for every action, there is an equal and opposite reaction. This means that when one object exerts a force on another object, the second object exerts an equal but opposite force on the first object. In the case of the three blocks on a frictionless surface, when m1 exerts a force on m2, m2 also exerts an equal but opposite force on m1.

In terms of free body diagrams, it is important to include all the forces acting on an object, including the force exerted by other objects on that object. This means that in the FBD for m1, you would include the force exerted by m2 on m1 as well as the applied force, weight, and normal force. Similarly, in the FBD for m2, you would include the force exerted by m1 on m2 as well as the normal force.

Now, let's look at the net force on block 1. Since the surface is frictionless, the only forces acting on m1 are the applied force and the force exerted by m2 on m1. Using Newton's second law, Fnet = ma, we can calculate the net force on m1 as 17.9N - 3.14N = 14.76N. This net force is what causes m1 to accelerate with an acceleration of 1.95m/s/s.

So why is the force exerted by m2 on m1 not included in the net force calculation? This is because this force is an internal force within the system of the three blocks. Internal forces do not contribute to the overall motion of the system, so they are not included in the net force calculation. This is why the applied force is the only external force included in the calculation.

In conclusion, it is important to include all the forces, including the forces exerted by other objects, in a free body diagram. However, when calculating the net force and acceleration of a system, only external forces should be considered. I hope this explanation helps clarify your confusion about Newton's third law and free body diagrams.