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Add the following vectors using trigonometry (i.e. cosine and sine laws).a. 7 m/s...

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oyechalphut | Student, Grade 10 | Honors

Posted February 19, 2013 at 6:04 AM via web

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Add the following vectors using trigonometry (i.e. cosine and sine laws).
a. 7 m/s [N30°E] and 2 m/s [S17°E].

b. 9 N [S2°W] and 11 N [N31°W].

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mjripalda | High School Teacher | (Level 3) Educator

Posted February 19, 2013 at 11:51 AM (Answer #1)

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To add vectors ` v_1` and `v_2` , determine the x and y components of each.

Then, add the x components.

`v_x = v_(1x)+ v_(2x)`

Also, add the y components.

`v_y = v_(1y) + v_(2y)`

Applying the Pythagorean formula, the sum of vectors `v_1` and `v_2` is:

`v = sqrt((v_x)^2 +(v_y)^2)`

And the direction of the resultant is:

`theta = tan^(-1) v_y/v_x`

For the above problem, to determine the x and y component of 7 m/s [ N 30`^o` E], use a right triangle. Since the reference is the North, the triangle is:

Note that the green is the x component and orange is the y component. And the angle 30`^o` is located between the orange and blue line and it belongs to the first quadrant of a unit circle chart.

Base on this triangle, the x component of 7 m/s [N30^oE] is:

`sin 30^o = v_(1x)/7`

`v_(1x)=7sin30^o`

`v_(1x) = 3.5`

And the y component is,

`cos30^o=v_(1y)/7`

`v_(1y)=7 cos 30^o`

`v_(1y)=6.06`

Do the same steps for the second vector 2 m/s [S 17`^o` E]. Since the reference is the south,  its triangle will be:

In the triangle 17`^o` is the angle between the orange and the blue lines. And it is located at the fourth quadrant of the unit circle. This means that the x component is positive but the y-component is negative.

So the x component of 2 m/s [S17`^o` E] is:

`sin 17^o=v_(2x)/2`

`v_(2x) = 2sin17^o`

`v_(2x)=0.58`

And the y component is:

`cos17^o=(-v_(2y))/2`

`v_(2y)=-2cos17^o`

`v_(2y)=-1.91`

Next, add the x components of the two vectors.

`v_x= v_(1x)+v_(2x)= 3.5+0.58=4.08`

Also, add the y components.

`v_y=v_(1y)+v_(2y)=6.06+(-1.91)=4.15`

The resultant of the given vectors is:

`v=sqrt(v_x)^2+(v_y)^2)=sqrt(4.08^2+4.15^2)=15.54`

Since both `v_x` and `v_y` are positive, the resultant vector is located at the first quadrant.

And its direction is:

`theta= tan^(-1) (v_y)/(v_x)= tan^(-1)4.15/4.08=45.49^o`

Note that the theta here is the angle from the positive x-axis (also referred as East axis).

Hence, the sum of the given vectors is 15.54 m/s [E 45.49`^o` N].

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