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I am not able to understand the FBD given above.

enter image description here

Here is my understanding: According to Newton's 3ed law, If apply the force with the help of my legs to the ground, the ground pushes me back with an equal and opposite force which is the net reaction force. The force that I applied on the ground can be resolved into two components: the weight and a horizontal component that is causing the pushing effect to the left. Similarly, the net reaction force can be resolved into two components: weight and fs(static frictional force). My question is:

  1. Does the normal reaction force which is one of the components of the net reaction force get cancelled with the weight( which is in turn one of the components of the net force that I applied to the ground). If this would happen then, the components of the action-reaction pair are getting cancelled right? also does the force on road by you and the force of the road on you get cancelled too? (see figure)(I am guessing no as they act on different objects)

  2. How is fs being produced? also if fs is the action force, where is the reaction force (-fs). shouldn't fs=-fs? But in the free-body diagram, fs(force of the road on you)=force on road by you.

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    $\begingroup$ Please only ask one question per post $\endgroup$ Apr 12, 2022 at 15:21
  • $\begingroup$ The weight is not a component of the force you apply to the ground. $\endgroup$
    – nasu
    Apr 12, 2022 at 19:23
  • $\begingroup$ Then should I ignore it? $\endgroup$ Apr 13, 2022 at 3:03

2 Answers 2

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Does the normal reaction force which is one of the components of the net reaction force get cancelled with the weight( which is in turn one of the components of the net force that I applied to the ground). If this would happen then, the components of the action-reaction pair are getting cancelled right? also does the force on road by you and the force of the road on you get cancelled too? (see figure)(I am guessing no as they act on different objects)

As you say, the forces don't "cancel" because they act on different objects. Of course you can mathematically add the vectors, but there isn't much use in that physically.

But if we have two vectors that are equal and opposite, $\mathbf F_1=-\mathbf F_2$, then we know their components along each direction are equal and opposite too:

$$\mathbf F_1=-\mathbf F_2$$ $$f_{1,x}\hat i+f_{1,y}\hat j=-(f_{2,x}\hat i+f_{2,y}\hat j)$$

Collecting each component then we must have $$f_{1,x}=-f_{2,x}$$ $$f_{1,y}=-f_{2,y}$$

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Newtons 3rd law is a vector equation $\vec{F}_{12}$ = -$\vec{F}_{21}$, that is, the force on object 1 from object 2 is equal and opposite to the force on object 2 from object 1. And since it is a vector equation it means that it is true for all components of the vector. So:

  1. Yes, the vertical force upward on you exerted by the ground is equal and opposite to the vertical force downward on the ground exerted by you.
  2. In the force diagram, you correctly draw the red horizontal arrow equal in length and opposite to the horizontal blue arrow. Since they are in opposite directions, this reflects indeed that the vertical vectors are opposite and equal: $\vec{F}^{\mathrm{friction}}_\text{on you by ground}$ = -$\vec{F}^{\mathrm{friction}}_\text{on ground by you}$
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  • $\begingroup$ so, force on the ground by you is a frictional force? $\endgroup$ Apr 13, 2022 at 3:10
  • $\begingroup$ Yes, the horizontal component is a friction force between foot and ground. Static for a normal run, kinetic if you run on a slippery ground. $\endgroup$
    – Jasoba
    Apr 14, 2022 at 11:54

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