Components of A Resultant Force

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The discussion centers on the concept of resultant forces and vector resolution. A scenario is presented where a box is pulled with a 10 N force at a 40° angle, resulting in horizontal and vertical components of 7.7 N and 6.4 N, respectively. The key question is whether pulling the box with these two component forces separately would produce the same acceleration as the single force. It is concluded that both methods would indeed result in the same acceleration, as the forces acting in parallel and perpendicular directions remain consistent. This reinforces the principle that component vectors are equivalent to their total force.
Bashyboy
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The following problem is one I have contrived by my own ability, so if there is any error please let me know.

Say I have a box, with a chain attached at one of the edges of the box, and I apply 10 N pull at a 40° angle with the surface of the box. Now I know we sometimes resolve vectors into their components for mere convenience of the addition of vectors. So in this case, we have a horizontal component of 7.7 N, and vertical component of 6.4 N. My question is: if I to pull with 6.4 N vertically, and a friend pull with 7.7 N horizontally, would the box accelerate in the same way if I were to only one force of 10 N at a 40° angle with the surface of the box?
 
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Two component vectors ARE equivalent to their total... just like 6.4 plus 7.7 equals 14.1...in that case the vectors point in the same direction...and if they were in opposite directions 6.4 - 7.7 is equivalent to -1.3...[signs are arbitrary]
 
Wow, thanks!
 
Yes, the accelaration in both cases would be the same because in both cases, he forces acting in the parallel and perpendicular directions would be the same, and since there is no question of variable mass, so the forces will produce the same accelaration.
 
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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