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Basic Question about forces

  1. Dec 18, 2011 #1
    1. Okay so in a book it says that a normal force is the force a surface exerts on an object so there can be no net charge when the object is at rest (correct me if i am wrong). But when setting up a problem i found in a book, specifically Oman's "How to Solve Physics Problems" they gave me this:

    force1.jpg

    but shouldn't the normal force face..
    force2.jpg

    Sorry if this is a stupid question. I messed up in my introductionary physics course my first quarter of freshmen year. I realized the reason for my terrible performance was because i basically just memorized how to do certain problem, skipping out on the WHY i solve it a certain way. So in short, your going to see a lot of questions from me! :). I just can't accept the fact that people are either good at physics or not, I wanna give physics all that I got before i can assure myself that i just suck lol
     
  2. jcsd
  3. Dec 18, 2011 #2

    Doc Al

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    What the diagram is showing are the components of the weight of the block. So Fn is the normal component of the weight, not the normal force. The normal force isn't shown in the diagram.

    But you're right. If you showed the normal force on the block it would point outward from the surface.
     
  4. Dec 18, 2011 #3
    You are absolutely right, the normal force DOES face that way!
    I guess they only wanted to know the magnitude of the force and not the direction.
     
  5. Dec 18, 2011 #4
    Wow thanks guys!! Lol why don't books just say that? All it took was a sentence to clarify. Thanks again!
     
  6. Dec 18, 2011 #5

    Doc Al

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    I assume that's a typo. The normal force has nothing (directly) to do with charge.
     
  7. Dec 18, 2011 #6
    Aha yeah sorry thats what happens when you take chemistry and physics at the same time. My brains fried, time to take a breather X.X. Thanks Guys!
     
  8. Dec 18, 2011 #7

    PeterO

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    Normal is another word for perpendicular.

    The weight force has a parallel component, Fp in the first diagram, and a Normal component, FN in the first diagram.

    What you have added in diagram 2 is the Normal Reaction Force.
    It is a normal force, since it is perpendicular to the surface.
    It is also the reaction of the surface to the imposed Normal component of the weight. Newtons 3rd law, for every force acting there is an equal and opposite force reacting

    Your difficulty comes about due to the fact that you, like an incredible number of people, leave out that key work Reaction.

    That is why , when we start to have objects traveling around banked tracks, where the Normal component of the weight force, is not equal in magnitude to the Normal Reaction Force in the situation - that people become horribly, and needlessly, confused.
     
  9. Dec 18, 2011 #8

    Doc Al

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    You make a good point. Calling the support force the 'normal force' can be confusing, since 'normal' just specifies a direction. On the other hand, I don't like calling it the normal reaction force, as that can confuse folks just learning Newton's 3rd law. After all, all forces are reaction forces.
     
  10. Dec 18, 2011 #9

    PeterO

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    Our evidence here is that only about 3% of students understand Newton's 3rd law anyway - even students of teachers who understand the 3rd law - and that unfortunately is not all of them - so never had a difficulty with stressing this as a normal reaction force.

    I always liked to include that the Normal Reaction force was "just strong enough" to make what ever is happening happen.

    I also did the analysis of blocks on slopes without splitting the weight force into components, but instead concentrating on how stong the Normal Reaction force had to be. Too small would predict the mass "digging into" the slope. Too strong and the mass is predicted to leap off the slope. The reaction force was instead "just strong enough" to predict/explain why the mass slides down the slope [if no friction is involved].

    Also: I am not sure that my decision to press sideways on a wall constitutes a reaction force - though the wall pushing back on me certainly does.
     
  11. Dec 18, 2011 #10

    Doc Al

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    A better term than 'reaction force' is passive force. That gets at the distinction you are trying to make, I think. Active versus passive. Your pushing on the wall is an 'active' force, the walls pushing back is a passive force. They constitute an 'action/reaction' pair (though I don't like that archaic terminology). Remove the active force and the passive force goes away.

    When describing the forces acting on that block (in this example), I do think it's a good idea to specify the support force from the surface, which acts normal to the surface, and not simply refer to it as the 'normal force' without additional words to clarify what that force is.
     
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