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  1. #1
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    forces question

    A mass of 20kg is suspended from a ceiling by two lengths of rope that make angles of 30^o and 45^o with the ceiling. Determine the tension in each of the ropes.

    i drew the triangle and found out the other angles using triangle properties and ended up with

    T1/sin45=T2/sin60=196N/sin95

    T1=196(sin45)/sin95=139.1 N and T2=196N(sin60)/sin95=170.3N

    the book shows the answers to be 143.48N and 175.73N and i can't figure out how i went wrong
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  2. #2
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    Is that "95" a typo for "45"? If not, where did you get it?
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Is that "95" a typo for "45"? If not, where did you get it?
    it should actually be a 75, i made a typo i was working on a couple problems at once. i drew the trangle for the problem with the angles, and added 45 and 30 to get 75^o, its hard to exaplin without my diagram present, obviously, so i am just looking for a verification of the calculations as i cannot see what i did wrong
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  4. #4
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    Quote Originally Posted by william View Post
    A mass of 20kg is suspended from a ceiling by two lengths of rope that make angles of 30^o and 45^o with the ceiling. Determine the tension in each of the ropes.

    i drew the triangle and found out the other angles using triangle properties and ended up with

    T1/sin45=T2/sin60=196N/sin95

    T1=196(sin45)/sin95=139.1 N and T2=196N(sin60)/sin95=170.3N

    the book shows the answers to be 143.48N and 175.73N and i can't figure out how i went wrong
    Edit: turns out I get the book answers when I set g = 9.8. For future reference it would be an immense help if you could tell us what the book uses for g as some use 9.8, some 9.81


    First of all we need to know what value of g your book uses. I am using g = 9.81 \text {N kg}^{-1}

    This is a problem with resolving the forces horizontally and vertically.

    Also these are common angles and the values should be known

    \sin (30) = \frac{1}{2}, \cos (30) = \frac{\sqrt3}{2}

    \sin (45) = \cos (45) = \frac{\sqrt2}{2}


    Vertically


    Down: mg = 20g

    Up: T_1 \sin (30) + T_2 \sin (45)

    If the object is in equilibrium the resultant force is 0 and so these two are equal

    20g = \frac{T_1}{2} + \frac{T_2 \sqrt{2}}{2} = \frac{T_1 + T_2 \sqrt2}{2}

    T_1 + T_2\sqrt2 = 40g (eq1)



    Horizontally


    Left: T_1 \cos (30)
    Right: T_2 \cos (45)

    Since there is no horizontal movement these forces are equal

    T_1 \cos (30) = T_2 \cos (45)

    This is equal to T_1\sqrt3 = T_2 \sqrt2 (eq2)


    --------------------------------------------


    We now have two equations and two variables

    T_1 + T_2\sqrt2 = 40g (eq1)

    T_1\sqrt3 = T_2 \sqrt2 (eq2)


    Spoiler:
    T_2 = T_1\, \sqrt{\frac{3}{2}}

    T_1 + \sqrt{2} T_1\, \sqrt{\frac{3}{2}} = T_1 + T_1 \sqrt3 = 40g

    T_1 = \frac{40g}{1+\sqrt3} = 148.63 \text{ N}


    T_2 = \frac{40g}{1+\sqrt3} \cdot \sqrt{\frac{3}{2}} = 175.90 \text { N}


    My answers are relatively close to the book's.
    Last edited by e^(i*pi); February 20th 2010 at 11:53 AM. Reason: attached diagram
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