Double Atwood's Machine. In Fig. 5.85 masses m1 and m2 are...

Double Atwood's Machine. In Fig. 5.85 masses $m_{1}$ and $m_{2}$ are connected by a light string $A$ over a light, frictionless pulley $B .$ The axle of pulley $B$ is connected by a second light string $C$ over a second light, frictionless pulley $D$ to a mass $m_{3} .$ Pulley $D$ is suspended from the ceiling by an attachment to its axle. The system is released from rest. In terms of $m_{1}, m_{2}, m_{3},$ and $g,$ what are (a) the acceleration of block $m_{3} ;(b)$ the acceleration of pulley $B ;$ (c) the acceleration of block $m_{1} ;(d)$ the acceleration of block $m_{2}$ ; (e) the tension in string $A ;$ (f) the tension in string $C ?(g)$ What do your expressions give for the special case of $m_{1}=m_{2}$ and $m_{3}=$ $m_{1}+m_{2} ?$ Is this sensible? figure can't copy

Double Atwood's Machine. In Fig. 5.85 masses $m_{1}$ and $m_{2}$ are connected by a light string $A$ over a light, frictionless pulley $B .$ The axle of pulley $B$ is connected by a second light string $C$ over a second light, frictionless pulley $D$ to a mass $m_{3} .$ Pulley $D$ is suspended from the ceiling by an attachment to its axle. The system is released from rest. In terms of $m_{1}, m_{2}, m_{3},$ and $g,$ what are
(a) the acceleration of block $m_{3} ;(b)$ the acceleration of pulley $B ;$
(c) the acceleration of block $m_{1} ;(d)$ the acceleration of block $m_{2}$ ;
(e) the tension in string $A ;$ (f) the tension in string $C ?(g)$ What do your expressions give for the special case of $m_{1}=m_{2}$ and $m_{3}=$ $m_{1}+m_{2} ?$ Is this sensible?
figure can't copy

Key Concepts

Tension Analysis in Ideal Strings

Since the strings are assumed to be light (massless) and inextensible, and the pulleys are frictionless, the tension remains constant along each individual string. This simplification is essential for writing the force balance equations for each mass and for solving for accelerations and tensions consistently.

Atwood’s Machine Principles

The classic Atwood’s machine provides a foundation for understanding how two masses connected by a string over a pulley accelerate under gravity. In this double Atwood machine, the same principles are extended with additional masses and pulleys, emphasizing the importance of understanding the basic Atwood setup before tackling the compounded system.

Special Case Evaluation

Analyzing the system under specific conditions—such as setting two masses equal or relating the third mass to the others—provides a valuable check on the general solutions. Such evaluations can offer insights into the behavior of the system and confirm that the derived expressions yield sensible results in limiting or symmetric cases.

Free?Body Diagram Analysis

Drawing free?body diagrams for each mass and pulley is crucial in visualizing and identifying the forces acting on the system. These diagrams help in systematically breaking down the forces—such as tension and weight—and properly applying Newton’s laws to each part of the system.

Newton’s Second Law

This concept is pivotal in solving the problem since each mass undergoes acceleration under the net forces acting on it. By applying F = ma to each component of the system, including the blocks and pulleys, one can set up equations that relate the gravitational forces, tensions, and accelerations.

Constraints in Pulley Systems

In systems with massless, inextensible strings and frictionless pulleys, accelerations and displacements of the connected masses are interrelated by kinematic constraints. Recognizing these relationships allows for the formulation of additional equations that govern how the accelerations of the individual masses and pulleys are connected.

Transcript

00:02 So we have a diagram that looks this way.

00:16 This is a mass m.

00:19 Then at this end we have a mass less pulley, connecting to m1 and m.

00:49 The tensions are t .a, also t .a.

01:02 Tension here is tc.

01:15 This is pully b.

01:23 So this is a dynamic screen.

01:25 Question you're going to apply newton's second law of motion to each mass and to each pulley now you should know that acceleration of pulley b is negative of the acceleration of mass 3 so the rate at 3 this mass accelerates downwards is a rate at which this pulley accelerates upward since the string is inextensible but of course their directions differ this goes up this goes up this goes that hence the negative sign.

02:06 Also we know that, let's take ab to be the average acceleration of police, the masses 1, end.

02:25 And so we can say that a1 plus a2 is twice of their average acceleration.

02:36 And using this, we can say that is equal to manage to a3.

02:44 Now for a, there is no net force on the massless pulley and so we conclude that tc equals to t a b is the massless pulley okay so massless right so what we do now is to formulate three or five simultaneous equations from this information that we can give in the first is to know that m1g minus t a equals m1 a1 applying the weight of this pulley is m1g.

03:50 Let's take note, and the weight of this pulley is m2g...

Please give Ace some feedback