# Homework Help: Understanding Atwood's machine

1. Oct 4, 2011

### Crusaderking1

1. The problem statement, all variables and given/known data

Answers in bold.

I want to know if I'm getting the right ideas here, since I basically have the same answer for every question. Basically, when acceleration is 0, there is NEVER a net force, and when there is constant velocity, acceleration is 0. I feel like my explanations for number 4 are shoddy.

In number 4, I basically use the equation to justify my answer, what is the actual reason for there being a net force when there is acceleration and none for velocity.

Also, is Fnet in the equation Fnet=mg the same thing as ƩF in ƩF =ma?

Thanks.

1. What was the acceleration of the system in Part 1? What does this tell you about the velocity of the system? Is there a net force on the system? Explain how you arrived at your answer.

The acceleration was 0 m/s2. The velocity of the system is constant because at 0 m/s2 there is no acceleration. There is no net force on the object because the mass of object one and two are both equal, and thus causing no acceleration resulting in no net force.

3. Ideally, under what condition is the net force on the Atwood’s machine zero? Is it possible for m2 to be in motion if the net force is zero? Explain.

The net force is 0 N when both the masses of object one and two are equal because the velocity will be constant. The mass of object two can still be in motion as long as the mass of object one has the same mass, and velocity is constant. When velocity is constant, acceleration is 0 m/s2, and the net force will be 0 N.

4. Is a net force being applied to an object when it is moving to the left with a constant velocity of 7.50 m/s? Is a net force being applied to an object when it is moving to the left with a constant acceleration of 7.50 m/s2? Explain each of your answers.

When an object is moving left at a constant velocity, there is no acceleration. When the acceleration is absent, then the net force is also zero. Just like part 1, the slope of the velocity is zero, and when considering that the sum of all forces is equal to mass multiplied by acceleration, then the net force is zero due to zero acceleration. There is a net force when the object is moving at a constant acceleration. The sum of all the forces is always equal to mass multiplied by acceleration. Clearly, there is a mass and acceleration unlike in part 1 where the acceleration is zero. The net force is now the acceleration 7.50 m/s2 multiplied by the unknown mass which will result in a net force.

2. Relevant equations

3. The attempt at a solution

Thanks.

Last edited: Oct 5, 2011
2. Oct 5, 2011

### Crusaderking1

So I have the basics down?

3. Oct 5, 2011

### Staff: Mentor

You have the basics down. You might want to invoke the name of Newton here and there by referring to his laws of motion where appropriate. In particular, consider the first and second laws and what they say about constant velocity and acceleration and the forces involved.

4. Oct 5, 2011

### Crusaderking1

Thank you for the input, its always much appreciated.

5. Oct 5, 2011

### Crusaderking1

I just have one more question.

Last edited: Oct 5, 2011
6. Oct 5, 2011

### Crusaderking1

Why scientifically does having a greater mass difference(on Atwood's machine) yield better results for finding the accepted value of g(9.80)?

I honestly do not know. Would it be that the there is a faster velocity as the difference in mass increases therefore yielding more accurate results?

7. Oct 5, 2011

### Staff: Mentor

Think about things like limits of measurement accuracies, the relative magnitudes of things that might make the machine depart from the ideal, etc. Essentially, systematic errors that affect real-life equipment.

8. Oct 5, 2011

### Crusaderking1

Would it be that the shorter time that the m2 falling is more accurate because it is more like free fall compared to masses with less difference between them, because the more time it takes to reach the ground, the more affect the forces of tension may have on the object?

I know a source of error is that we were not told to measure the mass but accept the values anyway.

Last edited: Oct 5, 2011
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