Resolution of forces by calculation, Physics

Resolution of Forces by Calculation:

One vector was resolved into two mutually perpendicular components.

2489_solution by calculation.png

So if there are several vectors each can be resolved into two components.

e.g.    F1   in directionθ1,   gives   F1 sinθ1,  and  F1 cosθ1

          F2   in directionθ2,   gives   F2 sinθ2,  and  F2 cosθ2

          F3   in directionθ3,   gives   F3 sinθ3,  and  F3 cosθ3

          etc.
Once the components have been resolved, they can be added to give a total force in the Datum direction, and a total force perpendicular to the Datum.

1019_solution by calculation1.png

2458_solution by calculation2.png

These additions can be done laborio


usly 'by hand' but the modern scientific calculator renders this unnecessary.

Each vector should be entered and multiplied by the cosine of its direction and added consecutively to arrive at a total, F cosθ.

This procedure should be repeated, by multiplying each vector by the sine of its direction, and added consecutively to give F sinθ.

To calculate the Magnitude of the resultant,

          Add  (F sinθ )2  +  (F cosθ)2    (= F2)

And find the square root of the addition  (=F)

To calculate the Direction of the resultant,

          Divide  (F sinθ )  by  (F cosθ)   (= Tanθ)

and find the Angle (direction) that has that resultant,

923_solution by calculation3.png

Note that the values of sine and cosine take both positive and negative values, depending on the direction.

The calculator automatically takes account of this during the procedure.

The only occasion when ambiguity can arise is when finding the angle of the direction (there may be an error of 180º).  This can be resolved by inspection. Note the following:

675_solution by calculation4.png

With reference to the ambiguity of direction,

note that  1949_solution by calculation5.png =  (A)  gives the same angle (direction) as   Fsinθ/-FCosθ =  (B).  Thus, F sinθ  and F cosθ  have to be inspected to see which is negative.  Solution (A) or (B) can then be selected.

Similarly,   Fsinθ/FCosθ gives the same result as  -Fsinθ/-FCosθ                    . 

Again, inspect the values of F sin q and F cosθ to see whether both are positive or negative.

Posted Date: 9/14/2012 7:43:13 AM | Location : United States







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