Masses on strings and pulleys (iWTSE.org)

[TomK]

Did you try a simpler problem among those before attempting solving the one we have been discussing?
This one is basically half as complex and the site shows a detailed solution:
https://i-want-to-study-engineering.org/q/pulley_dynamics_2/

It is important that you can see that both, velocity and acceleration of those pulleys, must be different and the reason behing that fact.
One (top-left) only redirects the tension of the string, while the other (bottom-right) has certain mechanical advantage (whatever it loses in force it gains in displacement and vice-verse).

View attachment 272656
View attachment 272657

View attachment 272658

I managed to do the first pulley question, though I have a question about this second one you have mentioned.

Why do the equations of motion for masses A and B assume the acceleration direction to be down? Why does the answer come out wrong if you assume one mass goes up (for A: T - mg = maA) and the other mass goes down (for B: mg - 2T = maB)?

If you do it with 'T - mg = maA' (for mass A), you get aA = -2g/3.
If you do it with 'mg - T = maA' (for mass A), you get aA = 2g/5 (correct).

 

[haruspex]

I managed to do the first pulley question, though I have a question about this second one you have mentioned.

Why do the equations of motion for masses A and B assume the acceleration direction to be down? Why does the answer come out wrong if you assume one mass goes up (for A: T - mg = maA) and the other mass goes down (for B: mg - 2T = maB)?

If you do it with 'T - mg = maA' (for mass A), you get aA = -2g/3.
If you do it with 'mg - T = maA' (for mass A), you get aA = 2g/5 (correct).

Both approaches will work, but the kinematic equation relating a_A and a_B is different. With T - mg = maA and mg - 2T = maB, a_A = 2a_B; switching A to mg - T = maA makes a_A = -2a_B.

 

[TomK]

Both approaches will work, but the kinematic equation relating a_A and a_B is different. With T - mg = maA and mg - 2T = maB, a_A = 2a_B; switching A to mg - T = maA makes a_A = -2a_B.

I think I get what you're saying, but I'd like to clarify.

If you make the decision that both masses go down, do you then decide to change the sign in front of '2aB', since you know that, in reality, one will go up if the other goes down?

Is the reason for using 'aA = +2aB' when one mass goes up and the other goes down because you have chosen the correct direction in the equations of motion, so there's no need to flip the sign?

 

[haruspex]

If you make the decision that both masses go down

You don't really decide such a thing in advance. All you decide to do is to select which direction is positive for each displacement/velocity/acceleration/force. Having made those choices, you then write the equations in a consistent manner and the answer pops out. Some variables may come out with positive values, others negative.

In the present case, the choice is positive up for both accelerations or positive up for one and positive down for the other. Flipping the choice for one acceleration flips its sign in all equations in which it appears.

 

[TomK]

You don't really decide such a thing in advance. All you decide to do is to select which direction is positive for each displacement/velocity/acceleration/force. Having made those choices, you then write the equations in a consistent manner and the answer pops out. Some variables may come out with positive values, others negative.

In the present case, the choice is positive up for both accelerations or positive up for one and positive down for the other. Flipping the choice for one acceleration flips its sign in all equations in which it appears.

I was taught to write the equations of motion such that the directions moved by each object would be consistent (e.g. if one goes up, the other goes down). If an acceleration comes out negative for an object, it means that the direction of acceleration you predicted was actually opposite for said-object. I think this method causes problems for me.

Your way seems easier and less prone to mistakes. Is my interpretation below correct?

Assume down is 'positive', therefore up is 'negative'. For an object, any forces pointing down have '+' in-front, whereas forces pointing up have '-' in-front. Solve the simultaneous equations, and the polarity of your acceleration solutions will reveal if the acceleration of each object was in the up or down direction.

 

[haruspex]

I was taught to write the equations of motion such that the directions moved by each object would be consistent (e.g. if one goes up, the other goes down). If an acceleration comes out negative for an object, it means that the direction of acceleration you predicted was actually opposite for said-object. I think this method causes problems for me.

Your way seems easier and less prone to mistakes. Is my interpretation below correct?

Assume down is 'positive', therefore up is 'negative'. For an object, any forces pointing down have '+' in-front, whereas forces pointing up have '-' in-front. Solve the simultaneous equations, and the polarity of your acceleration solutions will reveal if the acceleration of each object was in the up or down direction.

Yes, those are the two common approaches. Both work as long as you are consistent.

 

[TomK]

Yes, those are the two common approaches. Both work as long as you are consistent.

Thank you. I have now finished all 15 IWTSE problems under the 'contraptions' category.

 

[Lnewqban]

Thank you. I have now finished all 15 IWTSE problems under the 'contraptions' category.

Great news, Tom!
Hope that you feel more confident now than when you started this thread.
We all think and understand things in different manners, and it is very important to find out which approach best works for you.
You can do it!