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6.2 Summary Questions. Question 4

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    • Thread Starter

    ''An elastic string is stretched horizontally between two fixed points 0.80m apart. Anobject of weight 4.0N is suspended from the midpoint of the string, causing the midpoint to drop a distance of 0.12m. Calculate:
    a) the angle of each part of the string to the vertical.
    b) the tension in each part of the string.''
    How does one go about on drawing a diagram of this situation, will a diagram even make it easier or perhaps it's not needed. Where are the two fixed points located? at the middle of the string? If so, does it mean it gets stretched further since an
    object of weight 4.0N is being suspended from the midpoint and causing it to drop by 0.12m?

    We start by drawing out a diagram of the forces - the fixed points are at either end of the string so a V shape is formed with the 4N at the middle.

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    To find a) the angle \theta of each part, simply use trigonometry where \sin\theta = \frac{o}{h} and use \sin^{-1} to find \theta.

    To find b) the tension T in each part, is another case of trigonometry. If a system is stationary - i.e. the string didn't snap - then we know all the forces in equilibrium, because there is no net movement.

    This means that each T and the 4N forces can form a closed triangle, because they must all cancel out. From there, we can split it into a right angled triangle and use trigonometry again to find the hypotenuse, T!
    • Thread Starter

    Thank you!
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