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Basic question regarding dynamics concepts?

  1. Sep 6, 2013 #1
    I just started Physics 1 this semester. Although the kinematics we covered so far were simple, our teacher didn't do a particularly good job at explaining dynamics. In fact, all 10 students I talked to were left clueless by his lecture.

    What we don't understand at this point is the concept of F=ma. Here are a couple examples of the concepts I and the other students don't understand:

    1) Say I am pushing a table across the floor at a constant velocity. Since the velocity is constant, acceleration would be 0 (please correct me if I am wrong), which means that F=0. However, as soon as I stop applying force to the table, the table stops moving, so clearly I am applying some type of "force" on it while I am pushing it. So, if what I am doing is not considered force because there was no acceleration involved while pushing it at a constant velocity, what would it be, and how do you differentiate between what I am doing and a force?

    2) From our text (Spiral Physics), it states that if you are holding onto a rope, the force of gravity pulling you down is equal to the force of the rope pulling you up. That concept I get. However, it also states that if you are climbing the rope, the two forces (gravity and the rope) are still equal? This has me very confused, as it takes far more energy to pull yourself up the rope than to hold you in place. In addition, I know each time you move upward, you are exerting more of some type of "force" on the rope than if you were stationary. So, can someone please explain what is wrong with my perspective that is keeping me from understanding this?

    If someone could answer those two questions, me and my classmates would greatly appreciate it. If there are any decent resources to self-teach this stuff, that would be great as well, as our instructor just isn't explaining this stuff in a way that makes sense. From what I've heard from other students, it just gets worse as the semester goes on.
     
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  3. Sep 6, 2013 #2

    arildno

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    "1) Say I am pushing a table across the floor at a constant velocity. Since the velocity is constant, acceleration would be 0 (please correct me if I am wrong), which means that F=0. However, as soon as I stop applying force to the table, the table stops moving, so clearly I am applying some type of "force" on it while I am pushing it. So, if what I am doing is not considered force because there was no acceleration involved while pushing it at a constant velocity, what would it be, and how do you differentiate between what I am doing and a force?

    But are YOU the only one exerting a force on the table here?

    F=ma implies that the net sum of forceS equals zero if there is no acceleration, not that there aren't any non-zero forces acting upon it!

    Got that?
     
  4. Sep 6, 2013 #3

    Doc Al

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    A better way of writing Newton's 2nd law is ƩF = ma. When the acceleration is zero (velocity is constant) you can conclude that the net force on the object is zero. So you are certainly exerting a force on it, but there must be other forces acting, such as friction, that make the net force = 0.

    In order to climb the rope, each time you give it a pull you are accelerating yourself a bit. So you have to pull slightly more than your weight. If you climb the rope at an average constant speed, then on average you are exerting a force on the rope equal to your weight.

    As far as the energy required to maintain a tension in your arms at various positions while pulling versus just hanging, that's more a question of biology than simple physics.

    A useful resource to zip through, that might help with some basic concepts, is: The Physics Classroom
     
  5. Sep 6, 2013 #4
    Ah, that makes a lot more sense than what our instructor explained to us, or I guess I should say "didn't" explain to us.

    Just to make sure I am understanding this clear, what you are saying is that if I am pushing the table at a constant velocity, the force that I am applying to it at that point is equal to the forces resisting the force I am applying (gravity, air resistance, etc.), correct?
     
  6. Sep 6, 2013 #5
    Between your explanation and the one before yours, I think it all makes sense to me now. You guys made more sense in two small posts than our instructor did in about 30 minutes of trying to explain it to us.

    Thank you for that link as well. I will be sure to take a look at it later this evening when I have the time.
     
  7. Sep 6, 2013 #6

    arildno

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    Exactly!

    Once you remove your own force from the balance sheet, the friction from the ground, for example, REMAINS as a force acting on the table, thereby decelerating it, since there is no longer any force acting upon the table working in the direction of the table's motion (that was the force YOU provided to the table!)

    Just a minor quibble with what you wrote:
    If the ground is horizontal, the gravity works at 90 degrees to the motion of the table, and does not affect the balance sheet of the horizontally acting forces determining the state of motion of the table. Instead, the effect of gravity, seeking to drag the table down into the ground is counterbalanced by the normal force exerted by the ground, directly upwards (presuming the force you exert on the table is strictly in the horizontal direction).

    In general, vertically acting forces are balanced against each other, and horizontally acting forces are balanced against each other, yielding two balance sheets in total, one for each direction of motion (forces that works in a skew direction are decomposed in their vertical and horizontal elements, affecting thereby both balance sheets)
     
    Last edited: Sep 6, 2013
  8. Sep 7, 2013 #7
    Ok, got it. Thanks again for this information, I really appreciate it.
     
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