What forces are acting on an object during a bounce?

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The discussion focuses on the forces acting on an object, specifically a wrench, during a bounce. Key forces identified include weight acting downwards and the normal force acting upwards, but the importance of illustrating their points of application on a free body diagram (FBD) is emphasized. The concept of impulse is introduced, explaining that it is the integral of force over time and results in a change in linear momentum. Additionally, the impact of these forces can create torque, influencing the object's rotation during the bounce. Understanding these dynamics is crucial for analyzing the overall motion of the wrench.
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Homework Statement
While I was placing a spanner on the bonnet of a car it slips out of my hand
and falls to the ground.

On this diagram can you show me (draw) vectors showing the
force(s) acting on the spanner at the
instant it hits the ground. Label the forces.
Relevant Equations
N/a
Screenshot 2019-08-07 at 7.54.28 AM.png
This is not a homework question. I merely am interested in what forces are acting on it.
 
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Even if it's not homework, you can still take a cut at the solution. Start with a free body diagram (FBD) of the wrench and show the force(s) on it just as it makes contact with the ground. Also, are you familiar with the concept of "impulse"?
 
@berkeman No, I didn't know " impulse is the integral of a force, F, over the time interval, t, for which it acts. Since force is a vector quantity, impulse is also a vector in the same direction. Impulse applied to an object produces an equivalent vector change in its linear momentum, also in the same direction." Got it now. I drew one. All the forces I got was Weight (mg) downwards and Fn (normal force upwards)
 
Yes, there are two forces one up and the other down but that is not enough. Where, on the spanner, did you draw them as acting?
 
I didn't, really.
 
Jeff97 said:
I didn't, really.
So on FBDs, it's important to show where the forces are acting, and how that relates to any torques on the object. In this case, the force of hitting the ground (the impulse over the short bounce time) will produce a torque that causes a rotation as part of the bounce. Also, if the wrench was rotating when it hit the ground, that will affect the overall motion.
 
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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