Resolve the 50-lb force into its components acting along... (a) the x and y axes

Question

Resolve the 50-lb force into its components acting along...

(a) the x and y axes a defined in the figure using of course x-y components
(b) the x and y' axes as defined in the figure (solve by setting up a parallelogram and applying the Law of sines and Law of cosines)
(c) your buddy at work is baffled over the fact that the x-components of the 50-lb force calculated in part (a) and in part (b) do not have the same magnitude even though the x-axis remains defined in the same direction. Provide an explanation of why the two x-components have different magnitudes.

I'd like to know how to do the problem rather than just the answers. So if you could just give me tips on how to solve it.

Explanation / Answer

The components for the x-y axes are simple enough: the x-component is 50cos(450) =35.36, and the y-component is also 50cos(450) =35.36.

let the components of the 50 lb force along x and y' be p and q.

there is a parallelogram with side lengths p(along x) and q (along y') with their diagonal, ie the resultant having length 50. The angle between p and 50 is 45 deg and the angle between q and 50 is 15 deg.

Considering any of the triangles formed by the two sides and the diagonal, we can write form the sines' rule, p/sin(15) = q/sin(45) = 50/sin(120)

=>p=14.93 and q=40.78

The reason that the components along x are different in each case is that it is the vector 'sum' of the components that must stay constant irrespective of the choice of the axes. The individual components do not matter.

Note- you can carry out the analysis done for the x-y' axes for the x-y axes too. The results will be the same.

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