Three forces are applied to an object: (34.3 N, 62.0 degrees); (56.2 N, 125 degrees); and (45.4 N, 251 degrees) resulting in a net force of

Answer 1

This problem is just checking if you understand how to break vectors down into their components (and put them back together to find the resultant).

1. First, I would recommend sketching this out, using the 3 forces as vectors coming out from one point.
2. Remember that angles here are measured counter-clockwise from the + x axis.
3. Force 1 will be in the first quadrant, force 2 in the 2nd quadrant, and force 3 in the 3rd.
4. Drawing each vector will help you check that you have the right sign (+/-) when you use formulas to break down the vectors into vertical and horizontal components.
5. This will also give you a rough idea of what kind of answer you'll get. Since 2 of the 3 forces are upwards, I'd expect the resultant force to be upwards. With similar logic, I'd expect the resultant to be pointed leftward.
6. Now, solve for the components of each force. Personally, I like to do all of the horizontal components at once, then do all of the vertical components.
7. Here, the horizontal component of each force can be found by F_x = F*cos(Θ)
8. This gives you 3 separate values (2 of which should be negative), which you can add up to find the total horizontal force. This will be the x component of your resultant force vector.
9. Now, do the same for the vertical components. Each can be found using F_y = F*sin(Θ)
10. Again, add them all up, which will give you the y component of the resultant force vector.
11. Finally, you can combine the x and y components that you just found into one big resultant vector.
12. The magnitude of the resultant force can be found using the Pythagorean theorem, a² + b² = c². The x and y components that you've found are acting perpendicular to each other, leaving the resultant that you're looking for as the hypotenuse.
13. Now you can apply the same formulas we used above, just in reverse.
14. F_y = F*sin(Θ). This time, we know F (you just found it), and we already had F_y, so we can just solve for theta.
15. CAREFUL: There are 2 solutions when using the arcsin and arccos functions.
16. I would draw a new diagram on an x-y axis using the F_x and F_y values that you found. This should make it clear if the angle you find is correct, since it has to be in the correct quadrant.
17. Personally, I don't like to mess around with big angles in these problems. I would just draw my little right triangle with the x and y components, find an angle formed between the resultant and an axis, and then add or subtract 90, 180, or 270 to get the overall angle that stars from the +x axis.
18. For example, if the right triangle I drew was in the 2nd quadrant (that is, F_x is negative and F_y is positive) with its base along the y-axis, then i would subtract this value from 180 degrees to find the overall angle.