Find the acceleration of the masses

  • Thread starter Thread starter Monsterboy
  • Start date Start date
  • Tags Tags
    Acceleration
AI Thread Summary
The discussion revolves around determining the acceleration of two masses, M and 2M, connected by a pulley system. Initially, the assumption that both masses have the same acceleration led to incorrect results, but considering the acceleration of mass 2M as half that of mass M yields the correct answers. The key insight is that when mass M moves down by a certain distance, mass 2M moves only half that distance due to the constraints of the pulley system. This relationship can be visualized by understanding that the string's total length remains constant, which necessitates a corresponding reduction in the distance moved by the pulley. Ultimately, the analysis confirms that the acceleration of mass 2M is indeed a/2 when mass M has an acceleration of a.
Monsterboy
Messages
304
Reaction score
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
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
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
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
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:
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}##?
 
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}##
 
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
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
Back
Top