Interaction force pairs

1. Jan 28, 2010

kbm

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?

2. Jan 28, 2010

tiny-tim

Welcome to PF!

Hi kbm! Welcome to PF!
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. Jan 28, 2010

kbm

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. Jan 28, 2010

tiny-tim

Yes.
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. Jan 28, 2010

kbm

Alright that makes sense. Thanks for the help.