Understanding Force Pairs in Accelerated Systems: Explained for Beginners

In summary, the conversation discusses the concept of force and acceleration in static equilibrium and when objects are being accelerated. It is explained that in static equilibrium, the force exerted on an object by its support is equal and opposite to the force exerted on the support by the object. However, when objects are being accelerated, it can be difficult to understand how the force pairs are equal. The example of a hand holding up a mass that is too heavy is used to explain that the force of the mass on the hand is less than its weight (mg) due to the hand's acceleration away from the mass. It is clarified that if the hand and mass were both stationary, the force of the mass on the hand would be equal to mg.
  • #1
kbm
7
0
This is going to be a very elementary question, but I am having a hard time wrapping my head around it for whatever reason.

In static equilibrium, like a book on a table, I have no problem seeing that the force exerted on the table by the book is equal and opposite to the force exerted on the book by the table.

But when things are being accelerated I am having a hard time grasping how the force pairs are equal.

For example, if a hand is holding up a mass that is too heavy for the person to hold, then the mass and the hand will both be accelerated downwards. The forces acting on the mass would be the force of gravity, acting down, and the pushing force of the hand, acting up. The force of gravity is greater than the pushing force so there is a net force downwards.

How is it that the hand exerts an equal and opposite force on the mass?

The forces on the hand would be the weight of the mass, acting downwards, the force of gravity on the hand, acting downwards, and the pushing force of the arm on the hand, acting upwards.

The force of the mass on the hand is equal to the weight of the mass (mg), and the force of the hand on the mass is equal to the pushing force of the arm. But if these were equal, then the pushing force of the arm would equal the weight, or force of gravity, on the mass, an then there would be no net force downwards on the mass.

What am I getting confused?
 
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  • #2
Welcome to PF!

Hi kbm! Welcome to PF! :smile:
kbm said:
The force of the mass on the hand is equal to the weight of the mass (mg) …

No, the force of the mass on the hand is less than mg (if you like, because the hand is accelerating away, so it isn't getting the full oomph from the mass), and the force of the hand on the mass is the same (less than mg), causing the mass to accelerate downwards.
 
  • #3
Thanks for the reply.

Alright so just for clarification: If the hand was holding the mass up at equilibrium, then the force of the mass on the hand would be equal to mg, correct?

But because the hand is moving downwards, the force of the mass on the hand is <mg. So in an FBD the forces acting on the mass would be (mg) acting downwards and (<mg) acting upwards, so a net force down. Similarly, the forces on the hand would be (<mg) acting downwards, the weight of the hand acting down, and a push force acting upwards, which must be still LESS than (<mg).

Why though, if the hand and the mass are still in direct contact and accelerating down at the same rate, is the force of the mass on the hand LESS than it would be if they were both stationary?
 
  • #4
kbm said:
Alright so just for clarification: If the hand was holding the mass up at equilibrium, then the force of the mass on the hand would be equal to mg, correct?

But because the hand is moving downwards, the force of the mass on the hand is <mg. So in an FBD the forces acting on the mass would be (mg) acting downwards and (<mg) acting upwards, so a net force down. Similarly, the forces on the hand would be (<mg) acting downwards, the weight of the hand acting down, and a push force acting upwards, which must be still LESS than (<mg).

Yes. :smile:
Why though, if the hand and the mass are still in direct contact and accelerating down at the same rate, is the force of the mass on the hand LESS than it would be if they were both stationary?

If they were free-falling, they would still be in direct contact and accelerating down at the same rate, but the forces between them would be zero, wouldn't they?

This is an intermediate stage.
 
  • #5
Alright that makes sense. Thanks for the help.
 

1. What is a force pair in an accelerated system?

A force pair in an accelerated system refers to a pair of equal and opposite forces that act on two different objects or surfaces in contact with each other. These forces are responsible for the acceleration of the system and follow Newton's Third Law of Motion.

2. How do force pairs affect the motion of an object?

Force pairs cause an object to accelerate by exerting equal and opposite forces on it. This means that for every action force, there is an equal and opposite reaction force. These forces act on different objects, causing them to move in opposite directions.

3. Can force pairs cancel each other out?

No, force pairs cannot cancel each other out. According to Newton's Third Law, for every force, there is an equal and opposite force. This means that the two forces in a force pair cannot cancel each other out, but they can result in a net force of zero if they act on the same object.

4. How do force pairs contribute to an object's stability?

Force pairs contribute to an object's stability by keeping it in equilibrium. When equal and opposite forces act on an object, they cancel each other out, resulting in a net force of zero. This means that the object will remain in its current state of motion, whether it is at rest or moving at a constant velocity.

5. Can force pairs exist in non-accelerated systems?

Yes, force pairs can exist in non-accelerated systems. They follow Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. This law applies to all systems, whether they are accelerated or not.

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