Attachment 25634i 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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- Nov 10th 2012, 10:32 PMabdulrehmanshahMechanics new question just need a slight help
Attachment 25634i 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

- Nov 11th 2012, 07:50 AMskeeterRe: Mechanics new question just need a slight help
$\displaystyle T_1 \cos(40) = T_2 \cos(50)$

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

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

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

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

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

$\displaystyle T_2 = 192 \, N$

$\displaystyle T_1 = 161 \, N$

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

$\displaystyle T_2 \cos(50)$

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

$\displaystyle T_2 \sin(50) + T_2$

Resolving forces acting on the pulley ...

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

$\displaystyle \theta = \arctan\left[\frac{ T_2 \sin(50) + T_2}{T_2 \cos(50)}\right] = 70^\circ$ relative to the horizontal ... same as $\displaystyle 20^\circ$ relative to the vertical - Nov 11th 2012, 08:46 AMabdulrehmanshahRe: 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

- Nov 11th 2012, 08:54 AMskeeterRe: Mechanics new question just need a slight help