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Tension in a string of a object being pulled by a magnet

  1. Oct 16, 2007 #1

    ~christina~

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    Gold Member

    [SOLVED] Tension in a string of a object being pulled by a magnet

    1. The problem statement, all variables and given/known data
    An Iron bolt has a mass of 65.0g and hangs from a string 35.7cm ong. Top end of string is fixed. Without touching it, a magnet attracts the bolt so that it remains stationary displaced horizontally 28.0cm to the right from previously vertical line of string.

    a) draw free body diagram of bolt.

    b) find tension in string

    c.) find magnetic force on bolt.


    2. Relevant equations
    F= ma

    [tex]\sum F= 0[/tex]





    3. The attempt at a solution

    a.)
    [​IMG]

    I tried drawing the free body diagram and also in my opinion how the object would be in a drawing..

    I'm not sure what to do now since my text isn't clear on what I do to find tension and only has 1 example.

    Also I'm not sure about the magnet force pulling on the bolt but I assume it would be a F ...

    I do know that the sum of forces = 0 since it since it isn't moving...


    but after that I'm not sure what to do...

    ~I assume I find the angle that it goes out to the horizontal and then use that to find the x and y components of the tension??
    But I'm not sure how to resolve that either...


    so basically I'm stuck past a.)

    Thanks
     
  2. jcsd
  3. Oct 16, 2007 #2

    Astronuc

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    Staff: Mentor

    Resolve the tension into horizontal and vertical components.

    With the mass at rest, a = 0, so F = ma will not help here. On the other hand, [tex]\sum F= 0[/tex] will work, and in fact

    [tex]\sum F_x= 0[/tex] [tex]\sum F_y= 0[/tex]

    In order to find the angle of the string with the vertical, the string length length forms the radius of a circular arc (trajectory) of the bolt and the horizontal displacement would form the base of a triangle, with the other leg the radius - height that the bolt is displaced when moving upward in the arc.

    See what one can do with that.
     
  4. Oct 16, 2007 #3

    ~christina~

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    Well I did the problem and now I'm back with what I got...I refered to that site you linked me to the other time..hyperphysics

    let's see if I get this correctly...

    a.) tension in string (if I'm correct it is the m*g is the magnitude of the force in the down direction combined with the angle in y direction)

    Tx= 28.0cm (1m/10^2)= .28m
    Ty= 35.7cm(1m/10^2)= .357m

    sin(theta)= .28/.357= .7843
    sin^-1(.7843)= 51.66 deg

    T= mg/cos theta (y component of the force on string)
    m= 65.0g => 0.065kg
    g= 9.8m/s^2 (I think it isn't negative but...)

    T= [(.065kg)(9.8m/s^2)]/cos(51.66) = 1.027N => tension

    b.) the force of the magnet
    [tex]\sum Fx= T sin theta[/tex]

    T= 1.027N (assuming the tension force I found was fine above in part a)

    1.027N sin (51.66)= = .806 N (force of magnet on the bolt)


    Is this fine?

    Thanks :smile:
     
  5. Oct 17, 2007 #4

    Astronuc

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    Staff: Mentor

    The answers are correct.

    I am not sure about
    if one is writing T as tension. Tension is a force and therefore should have units of force, e.g. N.

    For a static situation, Ty = mg (weight) which points down, and Tx = Fmagnet, which acts horizontally.

    Using the angle with respect to vertical, Ty = T cos[itex]\theta[/itex] and Tx = T sin[itex]\theta[/itex].

    Given the horizontal displacement, 28 m, with a radius (hypotenuse) of 35.7 cm, then one can simply use sin[itex]\theta[/itex] = 28/35.7 and then cos[itex]\theta[/itex] = sqrt(1-(28/35.7)2).
     
  6. Oct 17, 2007 #5

    ~christina~

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    I don't understand why it's cos[itex]\theta[/itex] = sqrt(1-(28/35.7)2).[/QUOTE]


    I know that it relates to a circle but why the 1- ?? and the square root??


    could you explain it to me please...
     
  7. Oct 24, 2007 #6

    Doc Al

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    Astronuc is just using a famous trig identity that relates sines and cosines:

    [tex]\sin^2\theta + \cos^2\theta = 1[/tex]

    No need for that if you have a calculator.
     
  8. Oct 24, 2007 #7

    ~christina~

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    Oh..um Thanks Doc Al

    I was actually working on this again to practice for just in case there was a problem like this on my exam..

    I worked on it better though with actually writing sum Fx= F- Tsin theta= 0
    and etc..

    I did this problem the first time I tried to do a tension problem..hence the weirdness in my approach.

    Thanks for explaining :smile:
     
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