Find the resultant in unit-vector notation

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To find the resultant in unit-vector notation, the displacement vectors must be decomposed into their x and y components. The calculations yield Fx = 24.75 and Fy = 32.61, leading to a resultant magnitude of approximately 40.94 units. The direction of the resultant is found to be about 52.80 degrees from the +x axis. The final answers are 24.75 i hat + 32.61 j hat for part (a) and 52.9 units at 62.10 degrees for part (b).
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Homework Statement



Three displacement vectors of a croquet ball are shown in the figure, where |a vector| = 22.0 units, |b vector| = 25.0 units, and |c vector| = 10.0 units.

(a) find the resultant in unit-vector notation

____ i hat + ____ j hat

(b) Find the magnitude and direction of the resultant displacement.
____ units
____ ° (from the +x axis)

Homework Equations



http://img214.imageshack.us/img214/1280/physicsqj6.th.jpg http://g.imageshack.us/thpix.php



The Attempt at a Solution



I'm completely stumped. Can someone please teach me how to start this problem? Thank you.
 

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This is very similar to the last problem I helped you with. Decompose each vector to x and y components. Add 'em all up.
A unit vector has magnitude 1 so you will need to "normalize" your result. That is divide each component by the magnitude of the resultant vector.
 
how do I get the resultant in unit-vector notation?
 
can someone please check if my work and answer is correct?

|a vector| = 22* cos 90 = 0 ; 22* sin 90 = 22
|b vector| = 25*cos 45 = 17.68 ; 25*sin 45 = 17.68
|c vector| = 10*cos -45 = 7.07 ; 10*sin -45 = -7.07

Fx = 24.75 Fy = 32.61

24.75^2 + 32.61^2 = c^2
c = 40.94

tan^-1 (32.61/24.75) = 52.80 degrees

(a) 24.75 i hat + 46.75 j hat
(b) 52.9 units; 62.10 degrees
 
Last edited:
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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