Solving F = MA Problem: Confirm Solution & Extend Problem

  • Thread starter Knight526806
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In summary: You've got a complete free-body diagram for the entire system.This is a lot of work, and I don't think it's necessary. In fact, I would argue that it's actually more difficult to do it this way than just solving the problem without all the extraneous details. But if you want to be prepared for any questions about the problem, then I think it would be a good idea to at least understand how to do it this way.
  • #1
Knight526806
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So the Problem is with the picture which is attached.

Attempted Solution:
for m2: F = m2a = T - m2g
T = m2g
then for m1: F = m1a = T
a = T/m1
Then plug in: a = m2g/m1

then for all three blocks: F = (M = m1 = m2)a
so: F= (M +m1 +m2) (M2g/m1)

Now I'm pretty sure that's the right answer, but i need to know if my work is done correctly, every single part because I have to do this problem in front of the class :redface:. Moreover, how could I extend this problem. Like what would be a very related problem but make it a little more complicated and if you could provide a solution for the extension that would be awsome.
Thanks for all responses in advance.
 

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  • #2
Hello Knight526806,

Welcome to Physics Forums!

Knight526806 said:
then for all three blocks: F = (M = m1 = m2)a

I think you made a typo where you typed a couple of "=" signs instead of "+" signs. But...

so: F= (M +m1 +m2) (M2g/m1)

That sounds about right to me :approve: (Although I think you meant to type "m2" instead of "M2", but I'm just getting nitpicky now.)

If you want to be really clear, then you might want to break up respective forces into components. For example, when you started with m2, you might want to temporarily restrict your discussion of F to Fz to find the tension on the string. Similarly, if you wanted to, you could break up the x-direction force into the individual forces on the individual blocks and pulley. This includes the normal force of Block M to block m2, and the normal force from the pulley (the pulley is attached block M) to the string, which accelerates block m1 and the force needed to accelerate block M by itself. Then add up everything at the end (which will give the same answer that your already worked out above). You'll use Newton's third law a lot. This might seem like overkill, and in my opinion, it is sort of is overkill. But if you really want to be explicit, doing it this way should hold up better to questions.

Have phun presenting!
 
  • #3
Let me elaborate a little bit about what I meant with my last post.

The way you solved the problem is fine in my opinion. So there's nothing wrong with that. I'm not criticizing the way you solved the problem. But if you want to be completely prepared to answer any and all questions about the problem, there is a more detailed way to go about it. http://www.websmileys.com/sm/happy/535.gif

When you present your analysis, you might want to be at least prepared with some details in case somebody asks you about the forces on any individual block. Even it means just keeping this more detailed solution in your back pocket.

The way to go about doing this in explicit detail is to draw a free-body diagram of each individual block (you already know how they fit together, but you might want do also treat everything individually in case you get asked about it).

For example, consider block m2 on its own. Draw all the individual forces acting on block m2. You know that force m2g points down. There is a force T which points up. And you know that since block m2 does not accelerate in the z-direction, T = m2g (which you've already calculated). But wait, what other forces are on block m2? You know that it's accelerating at the rate a in the x-direction. So it is getting pushed by something. It's getting pushed by block M! And since you know the mass and acceleration of block m2, you can calculate the force on block m2 from block M.

And since block M is exerting a force on block m2, you also know that block m2 is exerting a equal and opposite force on block M (in the negative x-direction), due to Newton's third law of motion. So now you know two forces acting on block M, the force F (shown in the diagram) and the force that block m2 exerts on it (discussed above). (There may be other forces on block M too, that I haven't discussed above.)

What other forces are involved? Do the same procedure for block m1 and the pulley, eventually finding all the forces acting on block M. Once you have that, you can sum together all the forces acting on block M, and calculate what is the magnitude of force F (shown in the figure) that will cause block M to accelerate at a.

This is a more detailed way to work the problem (perhaps more detailed then really necessary), but it leaves no questions unanswered. Using this method, you should be able to handle any question that gets asked about the problem.
 

1. How do you confirm a solution to the F=MA problem?

To confirm a solution to the F=MA problem, you can use mathematical calculations and equations to show that the forces and accelerations in the problem are consistent with the given equation. You can also use experimental data and measurements to validate the solution.

2. What is the significance of solving the F=MA problem?

Solving the F=MA problem is essential in understanding the relationship between force, mass, and acceleration. It is a fundamental concept in physics and is used to explain the motion of objects in various scenarios, such as in everyday life or in more complex systems.

3. Can the F=MA problem be extended to more complex situations?

Yes, the F=MA problem can be extended to more complex situations, such as when multiple forces are acting on an object or when the mass of the object is changing. In these cases, the equation may need to be modified or combined with other equations to accurately represent the situation.

4. What are some real-world applications of the F=MA problem?

The F=MA problem has many real-world applications, including predicting the motion of objects in space, designing vehicles and machines, and understanding the behavior of fluids and gases. It is also used in fields such as sports and engineering to optimize performance and safety.

5. How can understanding the F=MA problem help in other areas of science?

Understanding the F=MA problem can help in other areas of science, as it is a fundamental concept that applies to many different phenomena. It can be used to explain the behavior of atoms and molecules, the movement of planets and stars, and the dynamics of biological systems. It also forms the basis for other important principles, such as Newton's laws of motion and the conservation of momentum.

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