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- anonymous

Two forces with magnitudes of 100 and 50 pounds act on an object at angles of 50° and 160° respectively. Find the direction and magnitude of these forces. Round to two decimal places in all intermediate steps and in your final answer.

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- anonymous

To determine the magnitude and direction of the forces, you must first resolve each into its horizontal and vertical vector components by using the cosine and sine functions. For the first vector, its horizontal component can be found via 100 lb cos 50 deg. (Make sure calculator is in degree mode). This yields 64.3 lbs in the positive x direction. For the vertical component we have 100 lb sin 50 deg which yields 76.6 lbs in the positive y direction.
For the 2nd vector's components we need to be mindful that 160 degrees is the same as 20 degrees above the negative x axis. Thus any x-values we obtain should be negative and our y-value will be positive. Using the same idea as before, 50 lb cos 160 deg yields -47.0 lbs in the negative x direction. To determine the y-value component, 50 lb sin 160 results in 17.1 lbs in the positive y direction.
Having resolved both vectors into components we now add up the horizontal and vertical components respectively to find a single component in each x and y direction.
Starting with the x components:
64.3 lbs + (-47.0) lbs =17.3 lbs in the positive x direction
Now the y components:
76.6 lbs + 17.1 lbs = 93.7 lbs in the positive y direction
Now, we use pythagorean theorem to determine the magnitude (resultant) of these vectors and then use the tangent function to determine the angle.
|dw:1358477633869:dw| (should be 17.3 not 17.1)
\[R = \sqrt{93.7^{2}+17.3^{2}} = 95.3 lb\]
To determine the angle/direction:\[\Theta =\tan^{-1} (\frac{ 93.7 }{ 17.3 }) = 79.5^{o}\]
Thus, your resulting vector will be 95.3 lb @ 79.5 degrees above the positive x-axis.
This is a valid answer and need not be converted into Newtons (N) as a pound is the English unit of measure of force. Good Luck!

- anonymous

That makes so much more sense. Thank you !!

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