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#### RobertoPink

##### New member

- Mar 27, 2018

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- Thread starter RobertoPink
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- Mar 27, 2018

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- Mar 1, 2012

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let the 150 lb force be directed along the positive x-axis and the 100 lb force be directed 40 deg relative to the positive x-axis in quadrant I (recommend you make a sketch)

using the method of components in the x and y directions, $R$ represents the resultant vector, $\theta$ is the resultant vector's direction relative to the positive x-axis ...

$R_x = 150 + 100\cos(40)$

$R_y = 0 + 100\sin(40)$

$|R| = \sqrt{R_x^2+R_y^2}$

$\theta = \arctan\left(\dfrac{R_y}{R_x}\right)$