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Math Help - Statics of particles in equilibirium

  1. #1
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    Statics of particles in equilibirium

    Hello all, i am currently working on statics of particles, and am having problems on the following question.

    What i done first was found the magnitude of the three angles using the cosine rule, i have both tried leaving the angles in exact form, and rounding off to the nearest decimal place, both giving me the wrong answer. The given angles are:
    ABC = 42
    BCA = 53
    CAB = 85

    What i do next was rearrange the vectors into a triangle of forces and apply the correct angles. I believe these are (after rearranging):
    ABC = 127
    BCA = 38
    CAB = 15

    maybe im wrong there, im not sure. After that i simply use the sine rule to find the tension in the two unknown strings, but i get the wrong answers >_<

    The answers are :
    T1 = 14.99 kg wt, T2 = 12.10 kg wt

    Can anyone see what i am doing wrong?

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  2. #2
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    Quote Originally Posted by jrp131191 View Post
    Hello all, i am currently working on statics of particles, and am having problems on the following question.

    What i done first was found the magnitude of the three angles using the cosine rule, i have both tried leaving the angles in exact form, and rounding off to the nearest decimal place, both giving me the wrong answer. The given angles are:
    ABC = 42
    BCA = 53
    CAB = 85

    What i do next was rearrange the vectors into a triangle of forces and apply the correct angles. I believe these are (after rearranging):
    ABC = 127
    BCA = 38
    CAB = 15

    maybe im wrong there, im not sure. After that i simply use the sine rule to find the tension in the two unknown strings, but i get the wrong answers >_<

    The answers are :
    T1 = 14.99 kg wt, T2 = 12.10 kg wt

    Can anyone see what i am doing wrong?

    note that the problem is unorthodox since kg's are not a unit of weight, but I will go with that since the solutions are in kg's.


    using the cosine law, angle between T_1 and T_2 is

    P = \arccos\left(\frac{10^2+12^2-15^2}{240}\right) \approx 85.46^\circ

    using the sine law, angle between T_2 and the horizontal is

    Q = \arcsin\left(\frac{10\sin{P}}{15}\right) \approx 41.65^\circ<br />

    angle between T_1 and the horizontal is

    R = 180 - P - Q \approx 52.89^\circ


    equilibrium in the x-direction ...

    T_1\cos{R} - T_2\cos{Q} = 0

    equilibrium in the y-direction ..

    T_1\sin{R} + T_2\sin{Q} = 20

    solve the system for T_1 and T_2 and you should get the desired results.
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