Two masses and two pulleys problem

In summary: So is it something like ##l = h + \pi r + (h - d_g)## where ##d_g## is the distance of the block from the ground. So if we differentiate twice we get that ##\ddot{d_g} = 2\ddot{h}##, which means that the rate at which the block accelerates is twice the rate at which the pulley...accelerates?In summary, we have a system of masses, strings, and pulleys with the constraint that the lengths of the strings remain constant. By applying Newton's second law to each object, we can derive relationships between the tensions in the strings and the accelerations of the masses. We also
  • #1
Mr Davis 97
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Homework Statement


Masses M1 and M2 are connected to a system of strings and pulleys as shown. The strings are massless and inextensible, and the pulleys are massless and frictionless. Find the acceleration of M1.

ATAeetS.jpg


Homework Equations


Newton's 2nd Law of motion

The Attempt at a Solution



So here is my line of reasoning. We have four objects with which we can use Newton's second law to derive relationships which will get us an explicit expression of the acceleration of block 1.

Newton's law for block 1:
##T_1 - m_1 g = m_1 a_1##

Newton's law for block 2:
##T_2 - m_2 g = m_2 a_2##

Newton's law for pulley with block 2:
##T_1 - 2T_2 = 0##
##T_1 = 2T_2##

Next, if we combine the equation for block 1 with the equation for block 2, and if we assume ##a_1 = a_2##, then we get,

##\displaystyle a_1 = \frac{g(2m_2 - m_1)}{m_1 - 2m_2}##

However, this is not the correct expression, because we must have that if M1 = M2 then ##\displaystyle a_1 = \frac{g}{5}##

Where am I going wrong with my reasoning?
 
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  • #2
Mr Davis 97 said:

Homework Statement


Masses M1 and M2 are connected to a system of strings and pulleys as shown. The strings are massless and inextensible, and the pulleys are massless and frictionless. Find the acceleration of M1.

ATAeetS.jpg


Homework Equations


Newton's 2nd Law of motion

The Attempt at a Solution



So here is my line of reasoning. We have four objects with which we can use Newton's second law to derive relationships which will get us an explicit expression of the acceleration of block 1.

Newton's law for block 1:
##T_1 - m_1 g = m_1 a_1##

Newton's law for block 2:
##T_2 - m_2 g = m_2 a_2##

Newton's law for pulley with block 2:
##T_1 - 2T_2 = 0##
##T_1 = 2T_2##

Next, if we combine the equation for block 1 with the equation for block 2, and if we assume ##a_1 = a_2##, then we get,

##\displaystyle a_1 = \frac{g(2m_2 - m_1)}{m_1 - 2m_2}##

However, this is not the correct expression, because we must have that if M1 = M2 then ##\displaystyle a_1 = \frac{g}{5}##

Where am I going wrong with my reasoning?
a1 is not equal to a2. Think: the moving pulley sinks by dx, how much does m2 move?
 
  • #3
ehild said:
a1 is not equal to a2. Think: the moving pulley sinks by dx, how much does m2 move?
But how does the second pulley move if it is massless?
 
  • #4
Mr Davis 97 said:
But how does the second pulley move if it is massless?
How does m2 move if the pulley does not move? The other end of the string is fixed to the ground.
 
  • #5
ehild said:
How does m2 move if the pulley does not move? The other end of the string is fixed to the ground.
Okay... But my equation ##T_1 - 2T_2 = 0## is correct since the mass of that pulley is zero?
 
  • #6
Mr Davis 97 said:
Okay... But my equation ##T_1 - 2T_2 = 0## is correct since the mass of that pulley is zero?
Yes.
 
  • #7
ehild said:
Yes.
Say for example that M1 is super heavy, and goes down. How does the second pulley accelerate upwards if it is massless?
 
  • #8
Why not? It pulls m2 upward.
 
  • #9
ehild said:
Why not? It pulls m2 upward.
But it's massless, so the net force on it is zero, which means there would be no acceleration it seems
 
  • #10
Mr Davis 97 said:
But it's massless, so the net force on it is zero, which means there would be no acceleration it seems
A massless object can accelerate without any force. F=ma. What can be a if both F and m are zero?
 
  • #11
ehild said:
A massless object can accelerate without any force. F=ma. What can be a if both F and m are zero?
The tension on string 1 is twice the tension on string 2, so would it be fair to suppose that the acceleration of mass 2 would be twice that of mass 1?
 
  • #12
Mr Davis 97 said:
The tension on string 2 is twice the tension on string 1, so would it be fair to suppose that the acceleration of mass 2 would be twice that of mass 1?
You get the relation between the accelerations from the constraints that the strings do not change their lengths.
 
  • #13
upload_2016-9-10_7-43-31.png
 
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  • #14
ehild said:
I'm just not seeing how to get the relation out of that... I know that I can say that the length of the string = so and so, and then differentiate twice... But I can't figure out the so and so.
 
  • #15
See the figure. If the pulley sinks by dx, the left piece of string becomes shorter by dx and the right one must become longer by dx. So the mass m2 moves downward by dx with respect to the pulley. But the pulley has moved downward by dx with respect to the ground, so the mass had to move by 2 dx downward with respect to the ground.
upload_2016-9-10_8-23-47.png
 

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  • #16
ehild said:
See the figure. If the pulley sinks by dx, the left piece of string becomes shorter by dx and the right one must become longer by dx. So the mass m2 moves downward by dx with respect to the pulley. But the pulley has moved downward by dx with respect to the ground, so the mass had to move by 2 dx downward with respect to the ground.
View attachment 105700
How would you derive this from the fact that the length of the string is constant? If we say that ##l = x_L + x_R + \pi r##, where r is the radius of the pulley, and the two x's are the left and right sections of the string, then if we differentiate twice we just get that ##\ddot{x}_L = - \ddot{x}_R##, which doesn't show that the rate at which the mass moves is twice the rate at which the pulley moves.
 
  • #17
The acceleration is with respect to the ground. You need to differentiate h twice, instead of xr.
 
  • #18
ehild said:
The acceleration is with respect to the ground. You need to differentiate h twice, instead of xr.
So is it something like ##l = h + \pi r + (h - d_g)## where ##d_g## is the distance of the block from the ground. So if we differentiate twice we get that ##\ddot{d_g} = 2\ddot{h}##, which means that the rate at which the block accelerates is twice the rate at which the pulley accelerates...?
 
  • #19
Mr Davis 97 said:
So is it something like ##l = h + \pi r + (h - d_g)## where ##d_g## is the distance of the block from the ground. So if we differentiate twice we get that ##\ddot{d_g} = 2\ddot{h}##, which means that the rate at which the block accelerates is twice the rate at which the pulley accelerates...?
Yes, and twice the acceleration of the other mass, only opposite direction.
 
  • #20
ehild said:
Yes, and twice the acceleration of the other mass, only opposite direction.
Do you get this from looking at the constant length of the other string for the other pulley?
 
  • #21
Mr Davis 97 said:
Do you get this from looking at the constant length of the other string for the other pulley?
Yes. Because of the length of the other string is also constant, and the left pulley is fixed, the accelerations of the hanging pulley and mass m1 are opposite and of equal magnitude.
 
  • #22
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  • #23
ehild said:
Yes. Because of the length of the other string is also constant, and the left pulley is fixed, the accelerations of the hanging pulley and mass m1 are opposite and of equal magnitude.
I have one more question about the nature of massless pulleys. If they are massless, wouldn't their acceleration be infinite according to the equation ##a = \frac{F}{m}##? How can we justify letting the masses of pulleys and ropes be zero?
 
  • #24
Mr Davis 97 said:
I have one more question about the nature of massless pulleys. If they are massless, wouldn't their acceleration be infinite according to the equation ##a = \frac{F}{m}##? How can we justify letting the masses of pulleys and ropes be zero?
That is why the net force on a massless pulley has to be zero. Otherwise the acceleration would be infinite. But we know that the pulley has finite acceleration. Therefore the net force is zero.
 
  • #25
ehild said:
That is why the net force on a massless pulley has to be zero. Otherwise the acceleration would be infinite. But we know that the pulley has finite acceleration. Therefore the net force is zero.
But if the pulley has a finite acceleration, how would the net force be zero?
 
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  • #26
Mr Davis 97 said:
But if the pulley has a finite acceleration, how would the net force be zero?
F=ma m=0 a= something. For example, a=5m/s2. F=0*5=0
 
  • #27
ehild said:
F=ma m=0 a= something. For example, a=5m/s2. F=0*5=0
But doesn't that contradict Newton's first law, which says that objects move at a constant velocity unless they have a net force? The pulley does not have a net force, yet it is accelerating...
 
  • #28
Zero mass is an approximation. Force wins over inertia of the bodies and makes them accelerate. The less the mass, the weaker is the force that products a certain acceleration. If a is constant,
F-->0 at the limit m-->0 according to Newton's second law.
 
  • #29
ehild said:
Zero mass is an approximation. Force wins over inertia of the bodies and makes them accelerate. The less the mass, the weaker is the force that products a certain acceleration. If a is constant,
F-->0 at the limit m-->0 according to Newton's second law.
Okay, so what if we wanted to analyze how the pulley accelerates using F = ma. If m = 0, how would we ever be able to find explicitly what a is?
 
  • #30
Mr Davis 97 said:
Okay, so what if we wanted to analyze how the pulley accelerates using F = ma. If m = 0, how would we ever be able to find explicitly what a is?
You cannot find the acceleration of a massless object using that equation. If an object is to be considered massless (which is never really true, of course) then the net force on it must be zero (else it would have infinite acceleration, which is not allowed) and its acceleration must be found by other constraints. In the case of a pulley, it just means it will rotate to match the linear acceleration of the the rope. Likewise, if the rope is massless then Its acceleration is to be found by the accelerations of the objects attached to it.
 
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  • #31
All right, I think I get it now. Back to the original problem.

So I have four pieces of information:

##T_1 - W_1 = m_1 a_1##
##T_2 - W_2 = m_2 a_2##
##T_1 = 2T_2##
##a_2 = 2 a_1##

When I combine all of this, I get the equation ##\displaystyle a_1 = \frac{g(2 m_2 - m_1)}{m_1 - 4 m_2}##, which is not correct because I need to have that if the masses are equal, then ##\displaystyle a_1 = \frac{g}{5}##. What am I doing wrong?
 
  • #32
Mr Davis 97 said:
All right, I think I get it now. Back to the original problem.

So I have four pieces of information:

##T_1 - W_1 = m_1 a_1##
##T_2 - W_2 = m_2 a_2##
##T_1 = 2T_2##
##a_2 = 2 a_1##

When I combine all of this, I get the equation ##\displaystyle a_1 = \frac{g(2 m_2 - m_1)}{m_1 - 4 m_2}##, which is not correct because I need to have that if the masses are equal, then ##\displaystyle a_1 = \frac{g}{5}##. What am I doing wrong?
You have to take the the sign of accelerations into account. m1 and m2 accelerate in opposite directions. If m1 moves upward, the hanging pulley moves downward, and so does m2. So a2=2a1 is not true.
 
  • #33
so a(2)=-2a(1)?
 
  • #34

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