Find the acceleration of the masses

In summary, the conversation discusses a problem involving pulleys and mass, where the acceleration of the mass 2M is found to be half of the acceleration of mass M. The conversation also touches on the concept of mechanical advantage and how it relates to force and displacement. The conversation also mentions the use of free-body diagrams and the importance of considering constraints in the problem. The conversation concludes with a discussion on visualizing the distances and positions involved in solving the problem.
  • #1
Monsterboy
303
96
Homework Statement
Consider the situation shown in figure. Both the pulleys and the string are light and all the surfaces are friction-less. Find the acceleration of the mass M and the tension in the string in the
figure given below.
Relevant Equations
Free-body diagrams and the forces acting on them.
1648611793049.png

I solved this problem by assuming that the acceleration ##a## is the same on the both the masses M and 2M but then answers were wrong, but if the acceleration of the mass 2M is considered as ##a/2## then I get the right answers, but I don't how exactly acceleration is getting halved for mass 2M.

How do I show that the acceleration of the mass 2M is ##a/2## if the acceleration of M is ##a## ?
 
Physics news on Phys.org
  • #2
Can you visualize pulley B as being a lever with pivot located at the lowest point, load located at the middle point and working force located at the highest point?
Then, combine that idea with the mechanical advantage concept: what you gain in force, you lose in displacement.

800px-Moveable-pulley-as-second-class-lever.svg.png
 
  • Like
Likes Monsterboy
  • #3
Monsterboy said:
Homework Statement:: Consider the situation shown in figure. Both the pulleys and the string are light and all the surfaces are friction-less. Find the acceleration of the mass M and the tension in the string in the
figure given below.
Relevant Equations:: Free-body diagrams and the forces acting on them.

How do I show that the acceleration of the mass 2M is a/2 if the acceleration of M is a ?
By using the fact that the string has constant total length.
Write an expression for the length of the string in terms of the locations of the masses and differentiate it wrt time a couple of times.
 
  • Like
Likes Monsterboy and Lnewqban
  • #4
Monsterboy said:
Homework Statement:: Consider the situation shown in figure. Both the pulleys and the string are light and all the surfaces are friction-less. Find the acceleration of the mass M and the tension in the string in the
figure given below.
Relevant Equations:: Free-body diagrams and the forces acting on them.

View attachment 299117
I solved this problem by assuming that the acceleration ##a## is the same on the both the masses M and 2M but then answers were wrong, but if the acceleration of the mass 2M is considered as ##a/2## then I get the right answers, but I don't how exactly acceleration is getting halved for mass 2M.

How do I show that the acceleration of the mass 2M is ##a/2## if the acceleration of M is ##a## ?
If you move mass ##M## by ##1## unit, then mass ##2M## moves by ##0.5## units. That's a constraint on the motion.

Note that if you pushed mass ##M## up, then mass ##2M## would not move. Again, in that case, an assumption of equal accelerations would be wrong.
 
  • Like
Likes Monsterboy
  • #5
haruspex said:
By using the fact that the string has constant total length.
Write an expression for the length of the string in terms of the locations of the masses and differentiate it wrt time a couple of times.
1648705595513.png

Ignoring the pulley radius.

##{x_0}## and ##{x_1}## are distances from mass 2M to pulley B, they are constants.
##{y_0}## and ##{y_1}## are the distances of mass M from pulley A.
##{l_0}## and ##{l_1}## are the distances between the pulleys A and B with respect to positions ##{x_0}## and ##{x_1}## of mass 2M and ##{y_0}## and ##{y_1}## of mass M respectively.Let the total length of the string be ##L##

##L = f(y, l)##

##L = 2{l_0} + {y_0}##
##L = 2{l_1} + {y_1}##

## L = 2l + y ##
Differentiating the above with respect to time
## \frac{dL}{dt} = 0 = \frac{2dl}{dt} + \frac{dy}{dt} ##
## \frac{2dl}{dt} + \frac{dy}{dt} = 0 ##
The above means ## 2({l_1} - {l_0}) = {y_1} - {y_0} ## in the time interval ##t## right ?
 
Last edited:
  • #6
That's essentially correct, but can you see directly that if the mass ##M## moves down by a distance ##\Delta y##, then the distance between the pulleys reduces by ##\frac {\Delta y}{2}##?
 
  • #7
PeroK said:
That's essentially correct, but can you see directly that if the mass ##M## moves down by a distance ##\Delta y##, then the distance between the pulleys reduces by ##\frac {\Delta y}{2}##?
Some people told me that but I am not able to visualize the distance moved by the pulley B to be exactly equal to ##\frac {\Delta y}{2}##
 
  • #8
Monsterboy said:
View attachment 299169
Ignoring the pulley radius.

##{x_0}## and ##{x_1}## are distances from mass 2M to pulley B, they are constants.
##{y_0}## and ##{y_1}## are the distances of mass M from pulley A.
##{l_0}## and ##{l_1}## are the distances between the pulleys A and B with respect to positions ##{x_0}## and ##{x_1}## of mass 2M and ##{y_0}## and ##{y_1}## of mass M respectively.Let the total length of the string be ##L##

##L = f(x, y, l)##

##L = 2{l_0} + {x_0} + {y_0}##
##L = 2{l_1} + {x_1} + {y_1}##

## L = 2l + x + y ##
Differentiating the above with respect to time
## \frac{dL}{dt} = 0 = \frac{2dl}{dt} + 0 + \frac{dy}{dt} ##
## \frac{2dl}{dt} + \frac{dy}{dt} = 0 ##
The above means ## 2({l_1} - {l_0}) = {y_1} - {y_0} ## in the time interval ##t## right ?
Yes, but it’s the accelerations you want, so differentiate a second time.
Monsterboy said:
Some people told me that but I am not able to visualize the distance moved by the pulley B to be exactly equal to ##\frac {\Delta y}{2}##
That's why I suggested this approach. You can use it in quite complicated arrangements without getting confused.
 
  • Like
Likes Monsterboy
  • #9
Monsterboy said:
Some people told me that but I am not able to visualize the distance moved by the pulley B to be exactly equal to ##\frac {\Delta y}{2}##
If you can't see something like this, then it's a lot of work and different variables to piece it all together. For eaxmple: the total length of the string, ##l##, is$$l = 2l_{AB} + l_{AM}$$And the position of pulley ##B## is the distance to ##A## from some fixed origin less the length of string ##l_{AB}##:
$$x_B = d_{OA} - l_{AB}$$And the vertical position of mass ##M## we can take to be ##y_M = l_{AM}##. Putting these together we get:
$$x_B = d_{OA} - \frac{l}{2} + \frac{y_M}{2}$$Finally, the distance from mass ##2M## to ##B## is constant, so:
$$x_{2M} = \frac{y_M}{2} + c$$for some constant ##c##, which we can take to be zero by a suitable choice of origin.

That's a lot of effort just to establish that $$x_{2M} = \frac{y_M}{2}$$which I would simply have written down as a system constraint. I wouldn't expect any exam marker to challenge that or expect more.
 
  • Like
Likes Lnewqban
  • #10
PeroK said:
That's essentially correct, but can you see directly that if the mass ##M## moves down by a distance ##\Delta y##, then the distance between the pulleys reduces by ##\frac {\Delta y}{2}##?
So half of ##\Delta y## of the string moves towards the pulley B from under it and the other half moves away, over it, summing up to ##\Delta y## length of string moving towards right, so the distance moved by the pulley is ##\frac {\Delta y}{2}## is this how you see it ?
 
  • Like
Likes Lnewqban
  • #11
Monsterboy said:
So half of ##\Delta y## of the string moves towards the pulley B from under it and the other half moves away, over it, summing up to ##\Delta y## length of string moving towards right, so the distance moved by the pulley is ##\frac {\Delta y}{2}## is this how you see it ?
Yes, the string runs twice between ##A## and ##B##. If ##B## moves ##1 \ cm## closer to ##A##, then you lose ##2 \ cm## of string.
 
  • #12
Monsterboy said:
Some people told me that but I am not able to visualize the distance moved by the pulley B to be exactly equal to ##\frac {\Delta y}{2}##
pully4a-gif.gif
 
  • Like
Likes nasu
  • #13
Lnewqban said:
Now that does lead to a complicated equation! Luckily the lower pulley magically shrinks as it rises so that the angle of the rope end doesn’t change.
 
  • Like
Likes SammyS

1. What is acceleration?

Acceleration is the rate of change of an object's velocity over time. It is a vector quantity, meaning it has both magnitude and direction. In simpler terms, acceleration measures how quickly an object is speeding up, slowing down, or changing direction.

2. How do you find the acceleration of masses?

To find the acceleration of masses, you need to know the mass of the object and the force acting upon it. You can then use the formula a = F/m, where a is acceleration, F is force, and m is mass. This formula is known as Newton's Second Law of Motion.

3. What is the difference between acceleration and velocity?

Velocity is the rate of change of an object's displacement over time, while acceleration is the rate of change of an object's velocity over time. In other words, velocity measures how fast an object is moving and in what direction, while acceleration measures how quickly the velocity of an object is changing.

4. Can you have negative acceleration?

Yes, an object can have negative acceleration. This means that the object is slowing down or changing direction in the opposite direction of its velocity. Negative acceleration is also known as deceleration or retardation.

5. How does mass affect acceleration?

The greater the mass of an object, the more force is needed to accelerate it. This is because the mass affects the inertia of the object, or its resistance to change in motion. Therefore, a larger mass will require a greater force to achieve the same acceleration as a smaller mass.

Similar threads

  • Introductory Physics Homework Help
Replies
23
Views
2K
  • Introductory Physics Homework Help
2
Replies
38
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
816
  • Introductory Physics Homework Help
Replies
10
Views
896
  • Introductory Physics Homework Help
Replies
12
Views
235
  • Introductory Physics Homework Help
Replies
4
Views
277
  • Introductory Physics Homework Help
Replies
2
Views
449
  • Introductory Physics Homework Help
Replies
7
Views
301
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
296
Back
Top