Solve Vector Problem: Find Magnitude & Angle of A+B+C

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The discussion focuses on solving for the magnitude and angle of the resultant vector A+B+C, given specific vectors A, B, and C with their respective magnitudes and directions. The user initially calculated the x and y components of each vector but made errors in the sine and cosine assignments for vector A. A participant pointed out that the x component should use the cosine function while the y component should use the sine function for vector A, correcting the user's approach. After receiving this guidance, the user felt confident that they could resolve the problem correctly. The thread emphasizes the importance of accurately converting polar coordinates to rectangular components in vector calculations.
mixedtape_15
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I can't figure this out for the life of me.

Vector A = 7.00 m and points 40.0o north of east. Vector B = 3.00 m and points 20.0o west of north, and Vector C = 3.00 m and points 35.0o west of south.
What is the magnitude of the resultant vector A+B+C?
What angle does the resultant vector A+B+C make with respect to the east?

So I've drawn it out and everything and I'm solving for "D" and what I drew out was some weird quadralateral(sp?). Anyways I solved for each of the Vectors and got the X Components and the Y Components. And this is what I got.

A -> Ax = (7.0m)(sin(40 deg)) = 4.5m
Ay = (7.0m)(cos(40 deg)) = 5.4m
B -> Bx = (3.0m)(cos(20 deg)) = 2.8m
By = (3.0m)(sin(20 deg)) = 1.0m
C -> Cx = (3.0m)(sin(35 deg)) = 1.7m
Cy = (3.0m)(cos(35deg)) = 2.5m

So then I added all the x components and all the y components and got the x and y components for D.
Dx = 4.5m - 2.8m + 1.7m = 3.4 m
Dy = 5.4m + 1.0m + 2.5m = 8.9m

and then I used the pythagorean theorem to get D which would be
3.4^2 + 8.9^2 = D^2
D = 9.5m

and I thought that was my answer but its telling me it was wrong. So I didn't even attempt to get the angle because that will probably be wrong to.
So yeah can someone help me please o:)
 
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mixedtape_15 said:
A -> Ax = (7.0m)(sin(40 deg)) = 4.5m
Ay = (7.0m)(cos(40 deg)) = 5.4m
B -> Bx = (3.0m)(cos(20 deg)) = 2.8m
By = (3.0m)(sin(20 deg)) = 1.0m
C -> Cx = (3.0m)(sin(35 deg)) = 1.7m
Cy = (3.0m)(cos(35deg)) = 2.5m
Excellent job showing your work -- that helps a lot in our ability to help you.

I believe that you are just mixing up the N and E components a bit. You are using the correct method, of converting from the "polar" coordinates they give you into rectangular (E,N) or (x,y) components. Looking at your Ax,Ay conversion, I can see that you've got them backwards. Since they give you the vector A as pointing 40 degrees north of east, the x component will be the cos() term, and the y component will be the sin() term. Make sense?

Just run back through the equations and check the sin() and cos() terms. My guess is that you'll get it right.
 
Remember, you can solve this with a ruler and a protractor.
 
Thank you so much for the help. It really helped me to figure out where I messed up and after that it was pretty easy :D
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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