How Do You Calculate Tension and Acceleration in an Atwood Machine?

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In summary, the conversation is about a question in a workbook regarding an Atwood machine and its acceleration and tension. The given solution in the workbook has a few typos, such as a decimal point missing in 1.9a and a misplaced decimal in 19.6a. These typos can cause confusion and frustration when trying to follow the solution. It is important to double check calculations and units to avoid errors.
  • #1
kitkat87
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Homework Statement



In my workbook it gives an example and solution for the following question:

The following diagram is of an atwood machine, which is a machine consisting of two masses: a massless string and a frictionless pulley. Due to the unequal masses on either end of the string, the system of masses will accelerate. Determine the acceleration of these masses, as well as the tension in the string.

So this is the question above. and the workbook solution is this:

Given: M1 = 1.0 kg and M2 = 1.9 kg

Required: a=acceleration and Ft = tension force

(Let the direction of acceleration be the positive direction, that is, up on the left and down on the right)

For M1

Fnet = Ft - Fg
m1a = Ft - mg
(1.0 kg) a = Ft - (1.0 kg)(9.8 n/kg)
Note: we can ignore units temporarily for ease of calculation.
1.0a = Ft - 9.8

For M2

Fnet = Fg2 - Ft
m2a = m2g - Ft
1.9a = 18.6 - Ft

Again, we leave units out for ease of calculation

Notice that there are two unknowns in each of the equations. Acceleration will be the same for both objects, since they are both connected. The magnitude of tension is also equal throughout the string. Therefore, to solve for acceleration, you can solve for Ft in each equation, then equate them.

*******This is the part where I got confused*******
*******The workbook throws in a random number in the next part and I have no idea where it came from, nor is there any explanation as to why it's there******

(From my understanding, I'm supposed to drag down the final numbers from the equations above and solve for Ft
However this workbook solution decides to defy all laws of mathematics and play mind games. Listed below is the next part of the given answer)

M1
Ft = 1.0a + 9.8 (correct)

M2
Ft = 18.6 - 19a (where did this number come from? and what happened to the 1.9a?)

Equate them and solve for a:

18.6 - 19.6a = 1.0a + 9.8
2.9a = 8.8

So now I'm even more confused because I have no clue where 1.9 went, they've now added in a magical mystery decimal of 19.6 and then at the end the calculations don't even add up properly. How does the workbook get 18.6-19.6 = 2.9 ? This is really confusion and frustrating.

Can someone please help explain, in detail, what I'm missing here?

I didn't do any of these calculations. This is just a given example of a question and a solution in my workbook that I'm apparently supposed to learn from.

The question goes on even further but I'm not going to put the rest in until I figure out this step properly.

Your help is greatly appreciated (pulling hair out)
 
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  • #2
kitkat87 said:

Homework Statement



In my workbook it gives an example and solution for the following question:

The following diagram is of an atwood machine, which is a machine consisting of two masses: a massless string and a frictionless pulley. Due to the unequal masses on either end of the string, the system of masses will accelerate. Determine the acceleration of these masses, as well as the tension in the string.

So this is the question above. and the workbook solution is this:

Given: M1 = 1.0 kg and M2 = 1.9 kg

Required: a=acceleration and Ft = tension force

(Let the direction of acceleration be the positive direction, that is, up on the left and down on the right)

For M1

Fnet = Ft - Fg
m1a = Ft - mg
(1.0 kg) a = Ft - (1.0 kg)(9.8 n/kg)
Note: we can ignore units temporarily for ease of calculation.
1.0a = Ft - 9.8

For M2

Fnet = Fg2 - Ft
m2a = m2g - Ft
1.9a = 18.6 - Ft

Again, we leave units out for ease of calculation

Notice that there are two unknowns in each of the equations. Acceleration will be the same for both objects, since they are both connected. The magnitude of tension is also equal throughout the string. Therefore, to solve for acceleration, you can solve for Ft in each equation, then equate them.
these equations are correct
*******This is the part where I got confused*******
*******The workbook throws in a random number in the next part and I have no idea where it came from, nor is there any explanation as to why it's there******

(From my understanding, I'm supposed to drag down the final numbers from the equations above and solve for Ft
However this workbook solution decides to defy all laws of mathematics and play mind games. Listed below is the next part of the given answer)

M1
Ft = 1.0a + 9.8 (correct)

M2
Ft = 18.6 - 19a (where did this number come from? and what happened to the 1.9a?)
textbook slipped a decimal should be 1.9a not 19a
I

Equate them and solve for a:

18.6 - 19.6a = 1.0a + 9.8
2.9a = 8.8
So now I'm even more confused because I have no clue where 1.9 went, they've now added in a magical mystery decimal of 19.6 and then at the end the calculations don't even add up properly. How does the workbook get 18.6-19.6 = 2.9 ? This is really confusion and frustrating.

Can someone please help explain, in detail, what I'm missing here?
another textbook typo 19.6a should be 1.9a
I didn't do any of these calculations. This is just a given example of a question and a solution in my workbook that I'm apparently supposed to learn from.

The question goes on even further but I'm not going to put the rest in until I figure out this step properly.

Your help is greatly appreciated (pulling hair out)
Ouch.
 

Related to How Do You Calculate Tension and Acceleration in an Atwood Machine?

1. What is tension?

Tension is a force that occurs when an object is pulled or stretched. It can also be defined as the pulling force exerted by the ends of a stretched object, such as a rope or a spring.

2. How is tension related to connected objects?

When two or more objects are connected, tension is the force that is transmitted between them. For example, in a pulley system, the rope connecting the pulleys experiences tension as it is pulled in opposite directions by the weight of the objects on each end.

3. What factors affect tension in connected objects?

The tension in connected objects is affected by the force applied to the objects, the properties of the objects (such as their mass and elasticity), and the angle at which the objects are connected.

4. Can tension ever be greater than the applied force?

Yes, tension can be greater than the applied force in certain situations. This can occur when the objects are connected at an angle, causing the force to be split into horizontal and vertical components. The vertical component can add to the applied force, resulting in a greater tension in the connecting object.

5. How is tension calculated in a system of connected objects?

To calculate tension in a system of connected objects, the net force on each object must be considered. The tension in a connecting object can be determined by balancing the forces acting on it, using Newton's second law of motion (F=ma).

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