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[Equilibrium and Pendulums] Do I have to consider the Perpendicular force?

  1. Jun 26, 2011 #1
    1. The problem statement, all variables and given/known data

    A pendulum bob of mass 2.5kg is pulled aside by a horizontal rope. If the tension in the rope is 3.50N, what is the angle between the pendulum and the vertical?

    mp = 2.5 kg

    Tr = 3.50 N

    Fg = 2.5kg * 9.8m/s2

    2. Relevant equations

    ∑F = 0

    Pocket full of trig functions and laws.

    3. The attempt at a solution

    I've been over thinking this one like a moe'sucka.

    Imagining the pendulum, one may speculate that there are 3 forces acting on the bob: the force of Gravity, Tension on the pendulum rope, and the force of Tension on the horizontal rope.

    It's a pretty basic under those terms, and solving it reveals:

    θ = 8.1°

    However, I don't feel like I've solved this correctly. [Edited explanation here, was going to rewrite it but then came to a conclusion.]

    My biggest question is: Do I have to concern myself with the applied force of gravity (perpendicular to the neutral force) in this instance?

    If so, the question becomes much more stressing to solve. In this case, the tension in the horizontal rope becomes the horizontal factor of the perpendicular force.

    I've expended most of my "goto" methods for solving these kinds of questions and even got a little crazy with trig stuff but I can't find that angle if I assume that 3.5N is only the x factor of the perpendicular force being balanced.

    Any suggestions would be appreciated.
    Last edited: Jun 26, 2011
  2. jcsd
  3. Jun 26, 2011 #2

    Doc Al

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    Staff: Mentor

    I understand gravity, but what are 'Neutral' and 'Pulling'?


    I don't know what you mean by 'neutral force' or 'applied force of gravity'. There are three forces acting on the pendulum bob. What are they?

    Maybe you can define what you mean by your terms and restate your question.
  4. Jun 26, 2011 #3
    Yeah, I could've used better terms. I'll edit.

    Pulling: Force of tension on the horizontal rope.

    Neutral: The force commonly written as Fn. What's the proper name for this?
  5. Jun 26, 2011 #4

    Doc Al

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    Staff: Mentor

    OK, that's the tension in the horizontal rope.

    Do you mean Normal force? That only applies between surfaces. Not applicable here.

    Perhaps you are thinking of the tension in the cord of the pendulum? That's the third force acting on the pendulum bob.
  6. Jun 26, 2011 #5
    This force is equivalent to the Tension on the pendulum rope though, correct? Fn has been used frequently to represent it in my instructional material.

    And you're right, there are just 3 forces. For some reason I was imagining that I had to counter a force that's used when working with moving pendulums. Imagining the system as a mass hanging from 2 points corrected it for me (same thing, different wording).

    I partially blame the hour.
  7. Jun 26, 2011 #6

    Doc Al

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    Staff: Mentor

    I've never heard the tension in a pendulum rope referred to as a 'neutral' or 'normal' force. I'd just call it tension. (But it doesn't matter what symbol you use for it, as long as you know it's a tension force.)
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