Can someone explain work and force problems to me?

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SUMMARY

This discussion addresses two fundamental physics problems related to work and force. The first question clarifies that no work is done by static friction since it does not act over a distance. The second question confirms that multiple forces can do work simultaneously if their resultant force is non-zero, allowing for vector addition of forces. An example illustrates that if two forces of 4N and 5N act on an object over a distance of 1 meter, they collectively perform 9 Joules of work. However, if the resultant force is zero, no work is done.

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  • Understanding of static friction and its implications on work
  • Knowledge of vector addition of forces
  • Familiarity with the concept of resultant force
  • Basic principles of work-energy theorem
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  • Study the work-energy theorem in classical mechanics
  • Learn about vector decomposition and resultant forces
  • Explore examples of static vs. kinetic friction in physics problems
  • Investigate scenarios where multiple forces result in zero net work
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hi i need help on two of my homework problems one on work and the other on force
As you walk, there is a static frictional force between your shoes and the ground. Is any work done? Explain your answer.

If three objects exert forces on a body, can they all do work at the same time? Explain. (for this question I'm thinking the answer is yes but I'm not sure about it)

I can't figure them out, it would be great if someone can help me Thanks!:smile:
 
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In your first case, the answer is no because the static friction force does not act over a distance.

For your second case, if you have three concurrent forces (fancy word for three forces acting along the same line), then you can vectorially (fancy word for add each contribution in the x and y direction)add them into a single resultant force. So the answer is yes. Because this single resultant force will do work on the body. Therefore, each of the forces that make up this resultant must do a proportional amount of work to contribute to the total work done.

Maybe some numbers will help. If I push you with a force of 4N, and my friend also pushes you at the same point with 5N for 1 meter, together we BOTH push you with 9N. You could say that I do 4Joules of work, and my friend does 5Joules of work. And together, we both do 9Joules.

Equivalently, I could say that we both impart a 9N force for 1 meter, and the net result is 9Joules. In the end, 9Joules is 9Joules.

You could expand this example to (n) number of friends pushing on you, provided you have that many friends :wink:
 
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thanks for the help cyrusabdollahi! Appreciate it!:smile:
 
Although the answer to the second question is yes, there can also be situations where three forces acting on a body does no work. This is when their resultant is 0, ie the three force vectors chould form the sides of a triangle taken in order .
 
When you push the ground backward with your foot, the ground exerts an equal force on you in the forward direction which propels you. For the time that you exert a force on the ground (and the ground exerts a force on you) the point of application of the force does not move hence the work done is zero.

As for your second question, in general if three (or more) forces act on a body so that their resultant is zero, all you can say is that the net force is zero

\vec{F_{net}} = 0

and (if the forces cause equal displacements)

\int_{i}^{f}\vec{F_{net}}\bullet d\vec{l} = 0

But in a complex case if force \vec{F_{i}} gives rise to a displacement \delta\vec{r_{i}} then the net work done is

\sum_{i}\vec{F_{i}}\bullet\delta\vec{r_{i}}

So the previous argument which seems obvious is valid for a rigid body where there can be no relative displacement between two points on the body.
 
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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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