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

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The discussion centers on calculating tension and acceleration in an Atwood machine with two unequal masses. The user expresses confusion over a workbook solution that introduces an incorrect value of 19a instead of the expected 1.9a, leading to discrepancies in the final calculations. The correct approach involves equating the tension equations derived from both masses, but the workbook's errors create frustration and hinder understanding. Clarification is sought on the source of these errors and how to properly derive the acceleration and tension. Accurate calculations are essential for mastering the concepts of dynamics in this context.
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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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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.
 
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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