# Action-reaction force diagrams

I have an issue with the force diagram examples of reaction forces.

It doesn't define where the "action"/applied force is coming from. If we were to draw free body diagrams, then for object 1 shouldn't it be force applied to the right and a reaction contact force against. For object 2 shouldn't it be a contact force to the right.

This leads to the overall question. Are forces measured through something or as an action from one object to the next? I know that if I exert a force, then that force should still be going through my arm. A question that whether this force applied should be in the free body diagram at all.

When we do free body diagrams for 2 or more objects, then the contact action/reaction pairs are not equal to force applied because the applied force external to the object?

• what are objects 1 and 2 in your first illustration of a person pushing a wall? (+1) for making the effort at improving the question – lineage Jun 15 at 1:17
• This might help - Why is the tension on both sides of an Atwood machine identical? – mmesser314 Jun 15 at 1:24
• In your second diagram are the objects initially at rest? – lineage Jun 15 at 1:58
• I already finished AP Physics 1. I was just thinking about where the force originates compared to what a problem shows. – Liberty Jun 15 at 2:04
• Yes, the second diagram is just a question of an external force compared to the force the arm exerts in the first. – Liberty Jun 15 at 2:05

To answer your question - forces are measured or as an action from one object to the next. Gravity is called 'action at a distance' but doesn't have to act through anything else.

The first diagram is badly drawn, that's true.

There is equal and opposite pairs of forces there. You can identify them by swapping words.

"the person pushes on the wall" (swap words) "the wall pushes on the person".

The diagram should have been drawn as a force on the person, to the left, starting at the person's hands and pointing left. Also an arrow to show the force on the wall, starting where the two arrows meet (on the original diagram) and pointing to the right. No other force is needed - for example a force in the persons arm.

In a Newton's 3rd law pair, the two forces act on different objects.

In the block question.

A person (let's say) pushes the 5kg block with 20N (swap) the 5kg block pushes the person with 20N.

Due to friction: The 5kg block pushes the floor to the right with up to 25N, (20N), (swap) the floor pushes the 5kg to the left with up to 25N (20N).

You get the idea, think it's best to stop there as there is no need for a contact force between the 5kg and 10kg block...

It doesn't define where the "action"/applied force is coming from

It, the diagram, doesn't have to. The source of the force isn't relevant to kinematics. How that force is generated, the internal processes of the force 'applier', the nature of the force itself are irrelevant to the act of balancing all forces acting on an object in free body diagrams (FBD). Only the value and direction of the force is needed.

But I think what you were trying to ask is more subtle and something else - the "where the force...is coming from". Do you mean, say in the first illustration, if the force is coming through the person's arms or whether the person's action force on the wall extends to some finite extent off of the wall perpendicularly, like a literally manifested red arrow?

If so, it isn't. From the perspective of a FBD, the action force is a vector acting $$at$$ the $$point$$ of contact. It has no spatial extent when we analyse it. Similarly, the normal reaction force of the wall on the person acts at exactly that point. These forces in fact exist only at that point when we analyse them.

As per Newton's third law they are vectorially equal and opposite. The FBD is

                  <---------.--------->
^      ^     ^
|      |     |
Reaction   |   Action
force     |   force
|
Pt. where force acts
(contact point)



fig 1 FBD of the point of contact between the wall and the person. To find the FBD of the person remove the action arrow and replace the dot with the person. For the FBD of the wall, remove the left arrow and replace the dot with the wall.

Are forces measured through something or as an action from one object to the next?

As explained above, no. We analyse forces at points where they act. When forces have spatial extent, they become force fields, but that may be an advanced idea for you at this stage.$$^1$$

I know that if I exert a force, then that force should still be going through my arm.

The force isn't going through your arm in so much as its being generated by the muscles of you arm.$$^2$$ Either way what pathway this so called "traveling force" takes through you arm to reach the point of application - the contact point with the wall - is irrelevant. What matters is that at that point, there is a force being applied. From the perspective of analysing force balance at that point, its extent and origin don't matter -it only exists at that point.

A question that whether this force applied should be in the free body diagram at all.

Of course, the FBD of the point of contact must include this force, as its being applied to it. Whether it was travelling down you arm or being generated by you r fingertips is not a criteria for exclusion/inclusion from the FBD. All that matters is, "Is this force acting at this point?"

then the contact action/reaction pairs forces are not equal to force applied because the applied force external to the object?

Not true. For e.g, in the case shown, if the left block be labeled $$1$$ and the right one labeled $$2$$, and the external $$10N$$ force be called $$F_e$$, with coeff. of static friction $$\mu$$, $$F_{ij}$$ represent the force on object $$i$$ by $$j$$, and the acceleration of the system be $$a$$ then for mass $$1$$, $$F_e-F_{12}-\mu m_1g=m_1 a ^{[3]} \tag{1}$$

If the external force isn't enough to overcome friction, and the masses stay at rest then $$a=0$$ and eqn. $$1$$ can be rearranged to show

$$F_{12}=F_e-\mu m_1g=-F_{21}$$

where the last equality comes from the third law. Since both friction and the external force are, well, external to mass $$1$$, action-reaction forces can in fact be equal to external forces.

$$^1$$ and the there is the concept of pressure but I recommend you clear this doubt first before studying that.

$$^2$$ In fact its the palm, no the arm, no the shoulder, no the torso, no the pelvis, no the legs, no the feet, no - the ground, yeah the ground you stand on that helps you push the wall. If the ground wasn't there you would only be able to push off of a wall (and in so doing push it)

$$^3$$ for mass $$2$$ its $$F_{21}-\mu m_2g=m_2 a$$

In this case if we push a wall it means we are pushing the wall it means we are applying force on the wall;similarly acc to Newton 3rd law the same force will be applied on us by the wall

things to be noted

• here there is a force on us by the wall and on wall by us So if we are drawing FBD of person then in horizontal direction there is only one force on the person which is by wall

• Action and reaction act on different bodies So there is only one force on person and one on wall Coming to "where the force is coming from" when we are saying a force we say it is by a mass A on massB I.e it itself says this force is coming from A Next "there is a force through your arm when you push" but it doesn't matter in FBD there is a force on the wall by person

• we say it by person not by arm Coz wall is a system and person as a whole is another.

1st thing-we should not say contact force but it is normal reaction Reason: contact force means force at the contact Friction-tangential component of contact force Normal rxn-the other component F²=f²+N² F-contact force remaining i guess u know

I think its answered and ignore if any grammar mistakes