Calculating Accelerations in an Atwood Machine with an Applied Force

In summary, the masses of the blocks are 20kg and 10kg, respectively. The blocks are initially at rest on the floor and are connected by a massless and frictionless pulley. An upward force, 124 N, is applied to the pulley. Block A experiences an acceleration of 9.81ms-2 while block B experiences an acceleration of 5.88ms-2.
  • #1
issacnewton
1,007
31

Homework Statement


The masses of blocks ##A## and ##B## are 20kg and 10 kg, respectively. The blocks are initially at rest on the floor and are connected by a massless and frictionless pulley. An upward force ##\vec{F}## is applied to the pulley. Find the accelerations ## \vec{a_A}## of block ##A##, and ## \vec{a_B}## of block ##B## when F is (a) 124 N ;(b) 294 N ; (c) 424 N.

Homework Equations


Newton's second law of motion. Relative acceleration formula

The Attempt at a Solution


I think it would be convenient to work in the pulley's frame. Since that is a non-inertial frame, I will need to introduce fictitious force in the analysis, so that I can use Newton's equations of motion as usual. Now let ##x_1## and ##x_2## be the coordinate of the blocks ##A## and ##B## as measured downward from the pulley. So, I am taking downward direction as positive. Since the system is accelerating upwards, I will need to introduce a fictitious force ##F## acting downward on each block. Let ##L## be the length of the string, then we have ##x_1+x_2=L##, which means ##\ddot{x_1} + \ddot{x_2} = 0##. Now equations of motion in the pulley's frame would be $$m_1g +F - T = m_1\ddot{x_1}$$ and $$m_2g +F - T = m_2\ddot{x_2}$$ where ##T## is the tension in the string and ##m_1## and ##m_2## are the masses of blocks ##A## and ##B## respectively. Subtracting two equations we get $$(m_1-m_2)g = m_1 \ddot{x_1} -m_2 \ddot{x_2} = m_1 \ddot{x_1} + m_2\ddot{x_1}$$ Solving, we get $$\ddot{x_1} = \frac{(m_1-m_2)g}{(m_1+m_2)}$$ Does it look alright ? There is no ##F## involved here. So I think I made mistakes somewhere.
1.PNG
 

Attachments

  • 1.PNG
    1.PNG
    3.8 KB · Views: 782
Physics news on Phys.org
  • #2
IssacNewton said:
Since the system is accelerating upwards, I will need to introduce a fictitious force ##F## acting downward on each block.

Fictitious force acting on the two masses will be different .
 
  • #3
If the force ##F## is applied externally, then usually we take fictitious force as ##F## acting in the opposite directions. So would it not be same for both the blocks ? Why is it different ?
 
  • #4
IssacNewton said:
If the force ##F## is applied externally, then usually we take fictitious force as ##F## acting in the opposite directions.

No .

Fictitious force is given by mass of the object times acceleration of the frame of reference , acting in the direction opposite to the accelerating frame .
 
  • #5
@IssacNewton this is a much simpler problem then you are thinking . Perhaps the moving pulley has thrown you off :wink: .
 
  • #6
I just wanted to solve it using the fictitious forces. So if the net force on the system of pulley and the blocks is ##F##, using we can take the total mass of the system, which is ##m_1+m_2##, so acceleration would be ##a = F/(m_1+m_2)##. So in that case, the fictitious force on ##m_1## would be ##Fm_1/(m_1+m_2)## and fictitious force on ##m_2## would be ##Fm_2/(m_1+m_2)##. Using this, we can modify the equations as follows
$$m_1g + \frac{Fm_1}{(m_1+m_2)} -T = m_1 \ddot{x_1}$$ and $$m_2g + \frac{Fm_2}{(m_1+m_2)} -T = m_2 \ddot{x_2}$$ Would this be ok ?
 
  • #7
IssacNewton said:
I just wanted to solve it using the fictitious forces. So if the net force on the system of pulley and the blocks is ##F##, using we can take the total mass of the system, which is ##m_1+m_2##, so acceleration would be ##a = F/(m_1+m_2)##

No . This is not the acceleration of the pulley ?:) .
 
  • #8
Well pulley is mass less, so how would we find its acceleration ?
 
  • #9
IssacNewton said:
Well pulley is mass less, so how would we find its acceleration ?

You can find pulley's acceleration after you calculate acceleration of the two masses :smile:

Have you solved this problem from ground frame ?
 
  • #10
No, I have not solved the problem from ground frame.
 
  • #11
IssacNewton said:
No, I have not solved the problem from ground frame.

Then please do that . Two steps and you get the answer :smile:
 
  • #12
Now if ##a_1## is the acceleration of block ##A## with respect to the ground, we also have ##F=2T## for the pulley. So equation of motion for block ##A## would be (upward directions are taken positive), ##T- m_1 g = m_1 a_1##. Since ##F=2T##, we have ##(F/2) - m_1 g = m_1 a_1##, so ##a_1 = \frac{F}{(2m_1)} - g##. And similarly, if ##a_2## is the acceleration of block ##B## with respect to the ground, we will get ##a_2 = \frac{F}{(2m_2)} - g##. Is it correct ?
 
  • #13
Yes .

Now , from the constraint relation you can find the acceleration of the pulley :smile: .

But since you are too eager to work from an accelerating frame ,would you like me to give you an exercise ?
 
  • #14
Yes, I would love that.
 
  • #15
OK .

Instead of force F acting on the pulley consider the pulley accelerating upwards with acceleration a = 10ms-2 and same masses as given in the OP , find the acceleration of the two masses ?
 
  • #16
Now using the relative acceleration, I can write the equation for the acceleration of the pulley(assume its ##a##).
##a_1 = a + \ddot{x_1}## and ##a_2 = a + \ddot{x_2}##. Adding the equations together, we get ##a_1 + a_2 = 2a + \ddot{x_1} + \ddot{x_2} = 2a##, since ##\ddot{x_1} + \ddot{x_2} = 0##. Solving for ##a##, we get
$$a = \frac{F}{4}\left(\frac{1}{m_1} + \frac{1}{m_2} \right) - g$$

Does it look right ?
 
  • #17
I have solved the acceleration of the pulley in post #16 for the original problem I posted. I will think about your new question in #15
 
  • #18
Now for your new problem in #15, let me use symbol ##a_3## for the acceleration of the pulley. So in the pulley's frame of reference, fictitious force of ##m_1a_3## will be acting downward on ##m_1## and fictitious force of ##m_2 a_3## will be acting downward on ##m_2##. So their equations of motion are
##m_1g + m_1a_3 - T = m_1\ddot{x_1}## and ##m_2g + m_2a_3 - T = m_2\ddot{x_2}##. Since ##F=2T##, we get the accelerations as $$\ddot{x_1} = g + a_3 - \frac{F}{2m_1}$$ and $$\ddot{x_2} = g + a_3 - \frac{F}{2m_2}$$ Does this look correct ?
 
  • #19
IssacNewton said:
Now using the relative acceleration, I can write the equation for the acceleration of the pulley(assume its ##a##).
##a_1 = a + \ddot{x_1}## and ##a_2 = a + \ddot{x_2}##. Adding the equations together, we get ##a_1 + a_2 = 2a + \ddot{x_1} + \ddot{x_2} = 2a##, since ##\ddot{x_1} + \ddot{x_2} = 0##. Solving for ##a##, we get
$$a = \frac{F}{4}\left(\frac{1}{m_1} + \frac{1}{m_2} \right) - g$$

Does it look right ?

Yes . But remember all what we have discussed so far is under the assumption that sufficient force is acting on the pulley such that both the masses are in air .

Go back and read the question again . You have been given three forces . Now think , why are three different forces given to you .
 
  • #20
IssacNewton said:
Now for your new problem in #15, let me use symbol ##a_3## for the acceleration of the pulley. So in the pulley's frame of reference, fictitious force of ##m_1a_3## will be acting downward on ##m_1## and fictitious force of ##m_2 a_3## will be acting downward on ##m_2##. So their equations of motion are
##m_1g + m_1a_3 - T = m_1\ddot{x_1}## and ##m_2g + m_2a_3 - T = m_2\ddot{x_2}##. Since ##F=2T##, we get the accelerations as $$\ddot{x_1} = g + a_3 - \frac{F}{2m_1}$$ and $$\ddot{x_2} = g + a_3 - \frac{F}{2m_2}$$ Does this look correct ?

No . F is not given to you :oldgrumpy: . You are only given acceleration of the pulley .Please read the question again in post#15 .

Let us discuss this part later after you have completed the original question .

Think about the three values of forces given to you .
 
  • #21
Ok. Let me solve for the original problem here. For ##F = 124 N## and ##m_1=20## kg, ##m_2 = 10## kg, using the formula I derived, we get ##a=-5.15 \,m/s^2##. Since ##a## is negative, it means that the pulley will not be lifted from the floor. For ##F = 294 N##, we get ##a = 1.23 \,m/s^2##, so here ##a>0## and pulley will be lifted with this acceleration. For ##F = 424 N##, we get ##a = 6.1 \,m/s^2##, so here again the pulley will be lifted with this acceleration from the ground. Does this look correct ?
 
  • #22
IssacNewton said:
Ok. Let me solve for the original problem here. For ##F = 124 N## and ##m_1=20## kg, ##m_2 = 10## kg, using the formula I derived, we get ##a=-5.15 \,m/s^2##. Since ##a## is negative, it means that the pulley will not be lifted from the floor. For ##F = 294 N##, we get ##a = 1.23 \,m/s^2##, so here ##a>0## and pulley will be lifted with this acceleration. For ##F = 424 N##, we get ##a = 6.1 \,m/s^2##, so here again the pulley will be lifted with this acceleration from the ground. Does this look correct ?

You are quite distracted by the acceleration of the massless pulley :cool: .

The original problem asks you to calculate accelerations of the two masses , not the acceleration of the pulley :mad:. You do not need acceleration of the pulley to determine acceleration of the two masses .It's the other way round :wink: .
Also note that expression for acceleration of the pulley in post#16 is valid only when the two masses are off the ground . This expression will not be valid to calculate acceleration of the pulley for the three different values of forces you have been given in the OP . Why ?

The important thing you might be overlooking/missing is that initially both the blocks rest on the ground .

To answer the original question , just analyse the forces on the two masses :smile:
 
  • #23
So when they are in contact with the ground, there will be weight downwards and normal reaction and the tension upwards. So, when the masses are lifted, their normal reaction would be zero. For the left mass, the normal reaction would be ##N_1= m_1 g - T = m_1 g - \frac{F}{2}## and for the right mass, the normal reaction would be ##N_2= m_2 g - T = m_2 g - \frac{F}{2}## when both of them are on ground. Now for ##F= 124\,N##, we get ##N_1 = 134 N## and ##N_2 = 36\,N##. Since in both cases, the normal reaction is positive, both the masses will remain on the ground. Now when ##F = 294\,N##, we get ##N_1 = 49\,N##, and ##N_2 = -49\, N##. Since ##N_2 < 0##, this means that the right mass will be lifted from the ground, but as ##N_1 > 0##, the left mass will remain on the ground. When ##F = 424\,N##, we get ##N_1 = -16\,,N## and ##N_2 = -114\,N##. Since both ##N_1 <0## and ##N_2 < 0##, both the masses will be lifted up from the floor and pulley will also move. So pulley will move when ##F = 294\,N## or ##F = 424\,N##.
Is this analysis correct ?
 
  • #24
IssacNewton said:
So when they are in contact with the ground, there will be weight downwards and normal reaction and the tension upwards. So, when the masses are lifted, their normal reaction would be zero. For the left mass, the normal reaction would be ##N_1= m_1 g - T = m_1 g - \frac{F}{2}## and for the right mass, the normal reaction would be ##N_2= m_2 g - T = m_2 g - \frac{F}{2}## when both of them are on ground. Now for ##F= 124\,N##, we get ##N_1 = 134 N## and ##N_2 = 36\,N##. Since in both cases, the normal reaction is positive, both the masses will remain on the ground. Now when ##F = 294\,N##, we get ##N_1 = 49\,N##, and ##N_2 = -49\, N##. Since ##N_2 < 0##, this means that the right mass will be lifted from the ground, but as ##N_1 > 0##, the left mass will remain on the ground. When ##F = 424\,N##, we get ##N_1 = -16\,,N## and ##N_2 = -114\,N##. Since both ##N_1 <0## and ##N_2 < 0##, both the masses will be lifted up from the floor and pulley will also move. So pulley will move when ##F = 294\,N## or ##F = 424\,N##.
Is this analysis correct ?

Looks alright .
 
  • #25
So when ##F = 294\,N##, the left mass will remain on the ground while the right one will start moving. Let ##a_2## be the acceleration of the right mass with respect to the ground. Then I have already found in post #12 that ##a_2 = \frac{F}{(2m_2)} - g##. Plugging the values, we get ##a_2 = 4.9\, m/s^2##

Now when ##F = 424\,N##, both the masses will be lifted from the floor, so using the formulae in post #12, we get ##a_1 = \frac{F}{(2m_1)} - g = 0.8\, m/s^2## and ##a_2 = \frac{F}{(2m_2)} - g = 11.4 \,m/s^2##.
I hope this is right. :smile:
 
  • #26
Yes
 
  • #27
Ok, now let me solve the problem given by you. ##a_3## is the acceleration of the pulley. Now in post #18, I got the equations of motion as ##m_1g + m_1a_3 - T = m_1\ddot{x_1}## and ##m_2g + m_2a_3 - T = m_2\ddot{x_2}##. We know that ##\ddot{x_1} + \ddot{x_2} = 0##, so we separate these two quantities from above equations. ##\ddot{x_1} = g + a_3 -\frac{T}{m_1}## and ##\ddot{x_2} = g + a_3 -\frac{T}{m_2}##. Adding them, we get $$\ddot{x_1} + \ddot{x_2} = 0 =2g+2a_3 - T\left( \frac{1}{m_1} + \frac{1}{m_2} \right)$$ Solving for ##T##, we get $$T = \frac{2m_1m_2 (g +a_3)}{(m_1 + m_2)} $$ Plugging ##T## back into the original equations we get $$\ddot{x_1} = g + a_3 - \frac{2m_2 (g +a_3)}{(m_1 + m_2)}$$ and $$\ddot{x_2} = g + a_3 - \frac{2m_1 (g +a_3)}{(m_1 + m_2)}$$ And now if ##a_1## and ##a_2## are the accelerations of the left and right block with respect to the ground, using the relative accelerations, we have ##a_1 = a_3 + \ddot{x_1}## and ##a_2 = a_3 + \ddot{x_2}##. Plugging the values of ##\ddot{x_1}## and ##\ddot{x_2}##, we get
$$a_1 = g + 2a_3 - \frac{2m_2 (g +a_3)}{(m_1 + m_2)}$$ and$$a_2 = g + 2a_3 - \frac{2m_1 (g +a_3)}{(m_1 + m_2)}$$

I hope this is correct...:oldbiggrin:
 

Related to Calculating Accelerations in an Atwood Machine with an Applied Force

1. What is an Accelerating Atwood machine?

An Accelerating Atwood machine is a physics apparatus that consists of two masses connected by a string or rope over a pulley. The pulley has negligible mass and friction, and the string is assumed to be massless. The machine is used to study the effects of acceleration, tension, and gravity on the system.

2. How does an Accelerating Atwood machine work?

The machine works by utilizing the force of gravity to accelerate the system. The heavier mass will experience a greater force of gravity, causing it to accelerate downwards. At the same time, the lighter mass will accelerate upwards due to the tension in the string. This creates an acceleration difference between the two masses, causing the system to move.

3. What are the applications of an Accelerating Atwood machine?

An Accelerating Atwood machine has various applications in physics, such as studying Newton's laws of motion, calculating acceleration due to gravity, and determining the coefficient of static and kinetic friction. It is also used in engineering to test the strength and durability of materials under different tensions and accelerations.

4. How is the acceleration of an Accelerating Atwood machine calculated?

The acceleration of the system can be calculated using the formula, a = (m1 - m2)g / (m1 + m2), where m1 is the mass of the heavier side, m2 is the mass of the lighter side, and g is the acceleration due to gravity. This formula assumes that the pulley has negligible mass and friction.

5. What factors affect the acceleration of an Accelerating Atwood machine?

The acceleration of the machine is affected by the mass of the two objects, the tension in the string, and the force of gravity. The acceleration will also depend on the friction and mass of the pulley, which can be minimized by using a low-friction pulley or assuming it to be massless. Additionally, air resistance and other external forces may also impact the acceleration of the system.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
289
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
30
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
751
  • Introductory Physics Homework Help
Replies
29
Views
1K
Replies
44
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
875
  • Introductory Physics Homework Help
Replies
6
Views
3K
Back
Top