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Math Help - Mechanics new question just need a slight help

  1. #1
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    Mechanics new question just need a slight help

    Mechanics new question just need a slight help-11111.gifi couldnt solve the last part where u have to find the resultant i know that the forumla to find the resultant but yet cant find it. I know that horizantal forces is 192cos50 but what would be the sum of vertical forces help
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    Re: Mechanics new question just need a slight help

    T_1 \cos(40) = T_2 \cos(50)

    T_1 = \frac{T_2 \cos(50)}{\cos(40)}


    T_1 \sin(40) + T_2 \sin(50) + 150 = 400

    T_1 \sin(40) + T_2 \sin(50) = 250


    \frac{T_2 \cos(50)\sin(40)}{\cos(40)} + T_2 \sin(50) = 250

    T_2\left[\cos(50)\tan(40) + \sin(50)] = 250

    T_2 = 192 \, N

    T_1 = 161 \, N


    forces on the pulley in the x-direction (leftward) ...

    T_2 \cos(50)

    forces on the pulley in the y-direction (downward) ...

    T_2 \sin(50) + T_2


    Resolving forces acting on the pulley ...

    |F| = \sqrt{[T_2 \cos(50)]^2 + [T_2 \sin(50) + T_2]^2} = 360 \, N

    \theta = \arctan\left[\frac{ T_2 \sin(50) + T_2}{T_2 \cos(50)}\right] = 70^\circ relative to the horizontal ... same as 20^\circ relative to the vertical
    Thanks from abdulrehmanshah
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    Re: Mechanics new question just need a slight help

    u deserve a million thanks well i just wanted to know what would be the vertical forces well thanks anyway u did a too much writing
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    Re: Mechanics new question just need a slight help

    Quote Originally Posted by abdulrehmanshah View Post
    u deserve a million thanks well i just wanted to know what would be the vertical forces well thanks anyway u did a too much writing
    Note that I have to do the problem from the start so that I understand how the entire problem develops.

    It's like coming into a movie after it has been running a while ... you really do not understand what is going on.
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