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Math Help - Forces question

  1. #1
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    Forces question

    I have another question that I can't make any sense of...

    A particle of mass 4kg is suspended from a point A on a vertical wall by a light inextensible string of length 130am.

    a) A horizontal force, P, is applied to the particleso that it is held in equilibrium 50cm from the wall. Find the value of P and the tension in the string.

    No problem with this part P = 16.4N and T= 42.5N

    b) By drawing a triangle of forces or otherwise, find the direction and magnitude of the minimum force that would hold the particle in this position.

    This bit has me stumped. Any hints would be much appreciated.



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  2. #2
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    Re: Forces question

    If you were to use a force upward rather than horizontally, it will require less force. Do problem (a) again assuming the force is at angle [itex]\theta[/itex] to the horizontal rather than horizontal and find the angle at which that force is aa minimum.
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  3. #3
    MHF Contributor ebaines's Avatar
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    Re: Forces question

    I think what they're asking is what happens if the force P was a general force applied to the particle, so that it has a vertical and horiziontal component. In part A the vertical componenet of P is zero, and you found that the horizontal component is equal to T\sin \theta. Now if you add in a vertical component of P, call it  P_y, then you have the following equations from the free body diagram:

    (1)  \sum F_y = 0 = T \cos \theta + P_y - W = 0

    (2)  \sum F_x = 0 = T \sin \theta - P_x = 0

    From (1) you have an expression for T that you can plug into (2) to find P_x as function of P_y. The magnitide of P is then:

    (3) |P| = \sqrt{P_x^2 + P_y^2}

    You can find the value for  P_y that gives a minimum for |P|by setting the derivative  \frac {d|P|}{dP_y} = 0.

    Post back with what you get for an answer.
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  4. #4
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    Re: Forces question

    Quote Originally Posted by ebaines View Post
    Post back with what you get for an answer.
    Ok I now realise what they wanted from the question.

    If you consider the three forces, the only one that is fixed (ie won't change) is W. So if you draw a triangle you can see that the shortest length for the P side will be when it is at a right angle to T. From this you can calculate the rest. My working and answer are attached as a jpeg.
    Forces question-mechforces.jpg

    I still feel like this hasn't completely settled in my mind (if that makes any sense) so if you have anything to add that might better my understanding of this please do!

    Thanks for your help
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  5. #5
    MHF Contributor ebaines's Avatar
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    Re: Forces question

    Yes, you are correct. For minimum P it is at right angles to T, and from the geometry you get P = (5/13)W and T = (12/13)W.
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