Exploring the Advantage of Using Pulleys in Two Block Problems

In summary, In the original Atwood machine problem, two masses are hanging from a pulley. The official answer says that the acceleration of the masses is ##\frac{1}{3}g##. However, my results show that the acceleration of the masses is ##\frac{1}{3}g+\frac{8}{3}g##. Thus, the tension in the rope must be providing more support than the official answer suggests.
  • #1
rudransh verma
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Homework Statement
Two blocks of mass m1=4kg and m2=2kg are hanging in Atwood machine. The total downward trust on the pulley is—
Relevant Equations
##F=ma##
##(T-m1g)-(T-m2g)=(m1+m2)a##
But I think this is wrong eqn.
 
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  • #2
Rather than pull an equation from the air, analyze each block separately. Then you can combine the two equations to solve for the tension in the rope.
 
  • #3
Doc Al said:
Rather than pull an equation from the air, analyze each block separately. Then you can combine the two equations to solve for the tension in the rope.
Are we not already analyzing exactly the same problem in https://www.physicsforums.com/threads/two-bodies-hanging-from-pulley.1011640/

The two masses there have a different ratio, but it is otherwise the identical situation.
 
  • #4
jbriggs444 said:
Are we not already analyzing exactly the same problem in https://www.physicsforums.com/threads/two-bodies-hanging-from-pulley.1011640/

The two masses there have a different ratio, but it is otherwise the identical situation.
Yes, the situations are identical, although the questions asked are slightly different. Here the problem asks for the downward force on the pulley (I assume) while the other thread asks for acceleration.

But you're right, it's essentially the same problem.

I'll leave this open for a bit, in case it was the wording of the problem that was throwing him. (Thanks for flagging this.)
 
  • #5
rudransh verma said:
But I think this is wrong eqn.
Why do you think it is the wrong equation? It was the correct equation in the other Atwood machine problem. As @Doc Al suggested, you need to find the tension in the rope. The equation you quoted here will not give it to you because the tension drops out as it is supposed to do.

In the other problem, you had a system of 2 equations and 2 unknowns, the tension and the acceleration. By eliminating the tension you got an equation involving the acceleration only. That's the equation you quoted here. Go back to the system of 2 equations in the previous problem and eliminate the acceleration to get an equation involving only the tension.
 
  • #6
rudransh verma said:
##(T-m1g)-(T-m2g)=(m1+m2)a##
But I think this is wrong eqn.
It is a correct equation, though perhaps not the best approach to the problem. In what sense do you think it is wrong? That it won't help answer the question or that it will produce the wrong answer? Those will depend on how you use the equation. What was your next step?
 
  • #7
kuruman said:
Why do you think it is the wrong equation? It was the correct equation in the other Atwood machine problem. As @Doc Al suggested, you need to find the tension in the rope. The equation you quoted here will not give it to you because the tension drops out as it is supposed to do.

In the other problem, you had a system of 2 equations and 2 unknowns, the tension and the acceleration. By eliminating the tension you got an equation involving the acceleration only. That's the equation you quoted here. Go back to the system of 2 equations in the previous problem and eliminate the acceleration to get an equation involving only the tension.
O yes!
 
  • #8
kuruman said:
In the other problem, you had a system of 2 equations and 2 unknowns, the tension and the acceleration. By eliminating the tension you got an equation involving the acceleration only. That's the equation you quoted here. Go back to the system of 2 equations in the previous problem and eliminate the acceleration to get an equation involving only the tension.
On taking g= 10 I got 2T= 53.34 N
On taking g= 9.8 I got 2T=52 N
Official answer is 54 N.
 

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  • #9
rudransh verma said:
On taking g= 10 I got 2T= 53.34 N
On taking g= 9.8 I got 2T=52 N
Official answer is 54 N.
Always show your work. However, your results fit with what I get, while the official answer does not.

The approach I took was to note that all the masses share the same speed. We have a net unbalanced gravitational force of 2mg. We have a total inertia of 6mg. So it is immediately clear that the acceleration of the masses will be ##\frac{1}{3}g##.

If the left hand (heavy) mass is accelerating down at ##\frac{1}{3}g## then tension must be supporting ##\frac{2}{3}## of its weight. That's ##\frac{8}{3}g## Newtons

If the right hand (light) mass is accelerating upward at ##\frac{1}{3}g## then tension must be providing ##\frac{4}{3}## of its weight. That's ##\frac{8}{3}g## Newtons

Sanity check: The two tensions match, as they must.

Total of the two tensions is ##\frac{16}{3}g## Newtons

To a reasonably appropriate number of significant digits...
With ##g=9.80665## I get ##52.3021## Newtons
With ##g=9.8## I get ##52## Newtons
With ##g=10.## I get ##53## Newtons.

I see no way to justify rounding ##53.\overline{3}## up to ##54## unless one has a deathly fear of odd numbers or a monstrously huge slab of depleted Uranium under the lab.
 
  • #10
jbriggs444 said:
I see no way to justify rounding 53.3― up to 54 unless one has a deathly fear of odd numbers or a monstrously huge slab of depleted Uranium under the lab.
I want to ask does using one pulley minimises the effort needed to pull a mass up. I don’t think this is true. ##F=m_1g+m_1a## and if we use a pulley then ##F-T+T-m_1g=m_1a##
 
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  • #11
rudransh verma said:
I want to ask why does we need more effort to pull a mass up directly as compared to using a pulley.
As I know Tension T required to pull up will be ##T=m_1g+m_1a##. But when we use a pulley the force will be ##(m_1+m_2)a+m_1g## to create the same acceleration a. So how is this force more than T?
Avoid introducing terms like ‘effort’ unless you define them. By ‘effort’, do you mean the net force on an object? Or the tension? Or something else?

Your first equation is OK: ##T=m_1g+m_1a##.

But the second equation isn't clear:
force = ##(m_1+m_2)a+m_1g##
You haven’t said which ‘force’ you mean.
And this equation ignores ##m_2##’s weight - there is no term ##m_2g##.

Maybe you can clarify the question.
 
  • #12
Steve4Physics said:
But the second equation isn't clear:
I was wrong with second eqn. I edited it. The force F required to pull mass m1 will be same ##m_1g+m_1a## whether we use a pulley or not. To take the advantage of pulley we need to add an extra weight. I hope I am right!
 
  • #13
rudransh verma said:
I was wrong with second eqn. I edited it. The force F required to pull mass m1 will be same ##m_1g+m_1a## whether we use a pulley or not.
The net force on ##m_1## is ##F_{net,1} = m_1a##. The tension in the rope is ##T =m_1g+m_1a##.
Take care to be absolutely clear to which of these you are referring, or it introduces confusion.

rudransh verma said:
To take the advantage of pulley we need to add an extra weight. I hope I am right!
It depends what you mean by 'advantage'.

The main advantage of a single pulley in practical use is that it changes the direction of the required force. It is often easier to raise a weight by pulling downwards rather than by applying an upwards force.

With an Atwood machine arrangement, you are raising a weight by allowing another weight (counterweight) to descend. This is rather like using a counterwight in a simple elevator:
https://www.researchgate.net/profile/Bolarinwa-Gabriel-Oladeji-2/publication/281288070/figure/fig2/AS:284481790464018@1444837152552/Arrangement-of-hoisting-rope-cabin-and-counterweight_Q320.jpg
 
  • #14
Steve4Physics said:
The net force on m1 is Fnet,1=m1a. The tension in the rope is T=m1g+m1a.
Take care to be absolutely clear to which of these you are referring, or it introduces confusion.
I am referring to T which is created by weight and acceleration. This will be the force necessary to lift m1.
Steve4Physics said:
It depends what you mean by 'advantage'.
I meant to reduce the necessary force to lift we need pulley and counter weight.
 

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What is a "two blocks and pulley problem"?

A "two blocks and pulley problem" is a common physics problem that involves two blocks connected by a rope or cable that is draped over a pulley. The blocks are typically on a horizontal surface and have different masses, and the goal is to determine the acceleration of the system and the tension in the rope.

What are the key concepts involved in solving a "two blocks and pulley problem"?

The key concepts involved in solving a "two blocks and pulley problem" are Newton's laws of motion, friction, and the relationship between force, mass, and acceleration. It is also important to understand how pulleys work and how to calculate the tension in a rope or cable.

What are the steps to solving a "two blocks and pulley problem"?

The steps to solving a "two blocks and pulley problem" are as follows:

  • 1. Draw a free body diagram for each block, showing all the forces acting on it.
  • 2. Apply Newton's second law of motion (F=ma) to each block to determine the net force acting on it.
  • 3. Use the relationship between force and acceleration to find the acceleration of the system.
  • 4. Use the acceleration to determine the tension in the rope or cable.
  • 5. Check your answer and make sure it makes sense.

What are some common mistakes made when solving a "two blocks and pulley problem"?

Some common mistakes made when solving a "two blocks and pulley problem" include forgetting to account for friction, not drawing accurate free body diagrams, and not correctly applying Newton's laws of motion. It is also important to pay attention to the direction of forces and accelerations, as they can affect the final answer.

How can I improve my understanding of "two blocks and pulley problems"?

To improve your understanding of "two blocks and pulley problems", you can practice solving different types of problems, work with a study group or tutor, and review the key concepts and equations involved. It can also be helpful to visualize the problem and think about real-life examples of pulleys and how they work. Additionally, seeking out additional resources such as textbooks, online tutorials, and practice quizzes can also aid in improving your understanding.

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