Here is the parallelogram rule in the file attached... Can someone please explain why the resultant force of these 2 vectors (30 N and 40 N) is 60 N? I know it's with an angle at 26ºC to the horizontal, so it wouldn't be 50 N. So how do you find this? How do you calculate the vectors when they are not straight lines shown. If someone is using this for an advantage of points, I'll report you. Thanks in advance :)

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Explanation:

The most straightforward way to compute the resultant of two vectors is to add their x- and y- components. In the attached, we have labeled the 30 N force "B" and the 40 N force "A". Their respective (x, y) components are ...

  A = 40(cos(0°), sin(0°)) = (40, 0)

  B = 30(cos(62.72°), sin(62.72°)) = (13.75, 26.66)

Then the components of the resultant are ...

  R = A + B = (40, 0) +(13.75, 26.66) = (53.75, 26.66)

Its magnitude is found from the Pythagorean theorem:

  |R| = √(53.75² +26.66²) = 60.00

Its direction is found from the components using the arctangent function:

  ∠R = arctan(26.66/53.75) = 26.38°

In a sort of shorthand notation, ...

  R = 60.0∠26.4°

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Alternate solution

In order to perform the above calculation, you need to know the angle between the vectors. The other angle in the parallelogram is the supplement of this. So, you can find the resultant by solving the triangle OAR, where O is the origin. Angle A in that triangle will be 180° -62.72° = 117.28°. This gives you enough information to use the Law of Cosines.

  OR² = OA² +AR² -2(OA)(AR)cos(A)

  OR² = 40² +30² -2·40·30·cos(117.28°) ≈ 3600.0

  |R| = √3600 = 60

Then you can find the angle AOR using the law of sines.

  sin(AOR)/30 = sin(117.28°)/60

  ∠AOR = arcsin(sin(117.28°)/2) = 26.38°

So, you now know ...

  R = 60.0∠26.4°