Still having problems with vectors

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Vector V1 has components V1x = -8.94 and V1y = 0, while Vector V2's components are incorrectly calculated; the correct components are V2x = 4.13cos(35°) and V2y = 4.13sin(35°). The user initially miscalculated V2's components, leading to incorrect results in part B. After recalculating, the sum of the x-components yields Vx = -5.56 and the y-component yields Vy = 2.37, resulting in a magnitude of approximately 6.04. The direction angle needs to be determined from the correct signs of Vx and Vy, which should be visualized in a diagram to find the angle counterclockwise from the +x axis.
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Vector V1 is 8.94 units long and points along the -x axis. Vector V2 is 4.13 units long and points at +35.0° to the +x axis.
(a) What are the x and y components of each vector?
V1x =
V1y =
V2x =
V2y =
(b) Determine the sum V1 + V2.
Magnitude
________
Direction
______° (counterclockwise from the +x axis is positive)

so i know that V1x=-8.94 and V1y = 0. i did the other components, but i got them wrong. i solved for V2y by 4.13cos(35) = 3.38. then solved for V2x by 4.13sin(35) = 2.37.

So for part B i did Vx = 8.94+2.37=11.31 and Vy=0+3.38=3.38. Then found the magnitude by V=the square root of (11.31^2) + (3.38^2) = 11.80. then for the direction i did tan (angle) = 3.38/11.31=.299. the inverse of that is 16.63 degrees.

Can anyone tell me what i did wrong
 
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confusedaboutphysics said:
so i know that V1x=-8.94 and V1y = 0. i did the other components, but i got them wrong. i solved for V2y by 4.13cos(35) = 3.38. then solved for V2x by 4.13sin(35) = 2.37.

So for part B i did Vx = 8.94+2.37=11.31 [...]

You lost the minus sign on -8.94. Minus signs count when you're adding components.
 
ohh ok thanks! but what about the V2x and V2y components. what did i do wrong? because i got those components wrong too so i can't do part B until i finish part A...help please!
 
Draw the triangle, with V2 as the hypotenuse.
Can you see now why V2y is not 4.13cos(35) but 4.13sin(35) ?
 
still having trouble. so here's what i got..but i got them wrong!

V1x= -8.94
V1y = 0
V2x= 4.13cos(35) = 3.38
V2y= 4.13sin(35) = 2.37

Vx= -8.94 +3.38 = -5.56
Vy= 0 + 2.37 = 2.37

V = [the square root of (-5.56)^2 +(2.37)^2] = 6.04

tan (angle)= 2.37/-5.56 = -.426

inverse tan (-.426) = -23.09

so what did i do wrong. I got the V1x & V1y correct, but everything else wrong. PLEASE HELP!
 
Your magnitude for V looks OK to me. What is it supposed to be?

As for the direction (angle) of V, draw a diagram that shows the x and y axes, and has Vx and Vy pointing in the proper directions, based on their signs. Look for a right triangle that has Vx and Vy as its two sides, and see where your angle of 23.09 degrees fits in (don't worry about the - sign here). From the diagram you should now be able to read off the angle that you were asked for, which is the angle measured counterclockwise from the +x axis to V.
 
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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