Your vectors can't be right, they're close(ish) but the inaccuracy may affect the result: most obviously 1^2 + 20^2 > 20^2



(jt20's were correct, but he mixed levels of accuracy, craig's were close, too, but zero is not probably not adequate) Math/physics/engineering rules - keep your accuracy (aka 'decimal places') consistent.



A= 28.68x + (-8.77)y

B= 1.04x + 19.97y



|A+B| = ((29.72x^2) + (11.20y^2))^-2 = 31.76



Theta = arctan 11.20/29.72 = 20.65 ccw from x.



The answer is fractions of a degree different from Craig's. How close do you need to be?



Cool java tool for visualizing here: http://www.surendranath.org/Applets/Math/VectorAddition/VectorAdditionApplet.html



Basic pic here: http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html

Convert to cartesian (i, j components)...easy with sines and cosines.

Add components, i+i, j+j

Convert back to magnitude and direction.

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Everyone... "m" was for meters, ya know the SI unit for length


where did you get those different values from? I posted the values from the question I was given.. so confused now


I see what you did.. with the Ax^2 + By^2 to get the result.. Why did you use those two values?


I can figure out why you used those two values for the arctan calculation after I know why you selected those two.. Thanks for your help everyone

Dude, it's a triangle.

Take a 2-D graph, X, Y coords.

Start at (0,0)
Put one vector end there, follow the magnitude out at the appropriate angle.
Affix the second vector to the tip of the previous.
The vector that connects the origin (0,0) to the tip of the second vector is the resultant (AKA A+B)

It's just a triangle...You can do trig, right???

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I used your vectors from post #6 and added them together to get the final vector, then calculated to magnitude and angle. Yak used the original problem statement and calculated more accurate vectors to get a better answer.

Quick and easy: go to the second link I posted. Read the block labelled "example", pay attention to the box with "component calculation" then scroll down to "Vector Addition, Two vectors". Plug the numbers in and you'll see.

Ax = 30 cos (-17) = 28.68
Ay = 30 sin (-17) = 8.77

Bx = 20 cos (87) = 1.04
By = 20 sin (87) = 19.97

You posted similar numbers for the vectors in your second post, but those are all rounded (i.e. 29, 9, 1, 20). You can use those values as components but you'll come up with a slightly different solution or the same one as Craig: 32m @ 20 degrees (sticking with no decimal places in the answer)

Close enough?

One of the great lines in the history of cinema.

Umm..you did recognize the "negative counterclockwise" or below the x axis and "counterclockwise" as above the x axis, I hope. That's why Ay is a negative value when adding the vectors.

It might be easier to see if you draw the A vector starting at 0,0; then draw the B vector stating at the tip of the A vector. The final vector goes from 0,0 to the tip of the B vector. You should be able to see how to apply the trig to the final triangle.

The java app in the first link does that. It also gives the x and y components. It doesn't explain how to calculate them, but it does make it easy to draw the pic.

they were intentionally left so he could decipher / decide what to do with them. sine of such a small angle is confusing when the cosine is equal to the full magnitude.

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ahh, I got it.. I see I see.. I understand what you did. I calculated it both ways and although the answer only needs 2 significant figures, anything more than 2 and the answer is different.

Thanks everyone