What is the Direction of Force in a Falling Object's Free-Body Diagram?

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In a free-body diagram for a falling object, the direction of force due to gravity is negative, indicating acceleration towards the ground at 9.8 m/s². The gravitational force acting on a 40 kg person is calculated as -392 N. There is a discussion about whether to represent this force as -392 N due to its direction or as a positive magnitude. It is clarified that force is a vector quantity, meaning it has both magnitude and direction, and the direction of movement does not necessarily align with the direction of the force. Understanding these concepts is crucial for accurately interpreting free-body diagrams in physics.
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Homework Statement



A 40kg person is falling at the speed of gravity, 9.8m/s^2. Wind resistance is not considered. On your free-body diagram, the direction they are falling (ground) is in the negative direction, hence gravity is causing an acceleration in the negative.

Homework Equations



F(gravity) = (40)(-9.8)
F(gravity) = -392N

The Attempt at a Solution



I am simply curious when solving a problem such as this, would N be a -392N since it is traveling in the negative on the free-body diagram, or is it seen a magnitude and considered a positive number?
 
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Ascendant78 said:
I am simply curious when solving a problem such as this, would N be a -392N since it is traveling in the negative on the free-body diagram, or is it seen a magnitude and considered a positive number?

Force is a vector therefore it has a direction. Also the direction of the movement of the object and the direction of the force do not have to me the same. For example a ball going up after you throw it moves up but the force acting on it is downward.
 
Thank you.
 
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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