How Does Newton's Third Law Apply When One Block Pushes Another?

  • Thread starter Thread starter aeromat
  • Start date Start date
  • Tags Tags
    Block
Click For Summary
In the discussion on Newton's Third Law, two blocks of mass m1 and m2 are analyzed under the influence of a constant force F applied to m1 on a frictionless surface. The key focus is on determining the acceleration of the system while considering the action and reaction forces between the blocks. It is clarified that the force exerted by m2 on m1 is equal in magnitude but opposite in direction to the force m1 exerts on m2. The problem simplifies the analysis by treating both blocks as a single unit for calculating acceleration, despite their differing masses. Understanding these forces is crucial for solving the problem accurately.
aeromat
Messages
113
Reaction score
0
Problem Type: Derivation

Homework Statement


2 blocks of mass m1 and m2 are placed in contact with each other on a smooth, horizontal surface as in the figure below. A constant force F is applied to m1 as shown.

(a) find the acceleration of the system


Homework Equations


Newton's Laws


The Attempt at a Solution


Picture of the problem:

F
-------> [ m1 ][m2]
=================================

Frictionless surface assumed.


FBD Sketch (are the action and reaction forces equal in magnitude?)

For m1:

Applied force -----> [ m1 ] <----- Reaction force by M2?
Fn and Fg are present.

For m2:

Applied force by m1-----> [m2]
Fn and Fg are present.

Is the reaction force received by m1 pushing m2 equal in magnitude but opposite in direction to the applied force on m1?
 
Physics news on Phys.org
Because the problem is only asking you the acceleration of the system, you can safely consider m1 and m2 the "same" block.

Remember, don't confuse yourself with free-body-diagrams. Usually only consider a force for calculation if it results in acceleration. In this case, the only force causing acceleration is the push (or friction, if it specifies).
 
But wait, the picture of my diagram states that the m1 mass is much bigger than the m2 mass. So, how would this work out? I am getting really confused with the action/reaction forces, and I don't know how I would work without them.
 
The force exerted by m2 on m1 is equal (in magnitude) but opposite (in direction) to the force m1 exerts on m2.

I would call this the reaction force.

I would call this the action force.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
23
Views
2K
Replies
2
Views
2K
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 8 ·
Replies
8
Views
11K
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
13
Views
3K