Question

The resultant of a certain system of forces has the X and Y components shown in Fig. 220. Determine the components of this resultant with respect to N and T axes rotated 30 deg counterclockwise relative to the X and Y axes.

Answer 1

First, you need to determine the resultant force and its angle. This one is real easy, since you only have one (x) and one (y) component. X = 300 and Y = 480 

1) sqrt (300^2 + 480^2) = ? (round your answer) 
2) tanθ = y/x Solving for θ: θ = tan^-1(y/x): θ = ? (round your answer) 

Now, you should have a resultant force value and its angle, from zero, in the first quadrant and between the Y and N axes. Draw yourself a diagram showing all of the axes, the resultant and the angles. You're going to have to do some very easy math to determine the angles between the Y and N axes and the resultant. Call these new angles α and β. α is the angle between N and the resultant and β is the angle between T and the resultant. Hint: one of them is 28 deg. 

Once you've gotten those figured, the Y and N axes become your new axis, ignore the X and Y. Find the X and Y components of the resultant (again). ? * cos(α) = 500 and ? * sin(β) = 266 (round your answers)

Answer 2

The resultant with respect to N and T axes rotated 30 deg counterclockwise relative to the X and Y axes is 566.03lb

Let the resultant force of the system be given as R. The resultant R with respect to the component is expressed as:

$N=\sqrt{R_x^2+R_y^2}$

Given the following parameters:

Ry = 480lb

Rx = 300lb

$N=\sqrt{480^2+300^2}\\N = \sqrt{320,400}\\N = 566.03lb$

Similarly; N = T = 566.03ib

Hence the resultant with respect to N and T axes rotated 30 deg counterclockwise relative to the X and Y axes is 566.03lb

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