System of pulleys - find acceleration

[blackboy]

Homework Statement

An object of mass m1 on a frictionless horizontal table is
connected to an object of mass m2 through a very light pulley
P1 and a light fixed pulley P2 as shown in Figure P5.34.
(a) If a1 and a2 are the accelerations of m1 and m2, respectively,
what is the relation between these accelerations? Express
(b) the tensions in the strings and (c) the accelerations
a1 and a2 in terms of the masses m1 and m2, and g.

Homework Equations

The Attempt at a Solution

How are these two pulleys connected? I don't think the string throught pulley 2 connects to m1 or else the tension and acceleration would be the same. I am having a hard time visualizing what is happening. Any help would be appreciated. Thank you.

 

[rl.bhat]

In all such problems we have to follow five steps.
Step 1: Find out the number of inextensible ropes. For every inextensible rope there will be one equation.
Step 2: For every inextensible rope, find the bodies that are connected to the rope.
Step 3: To relate the acceleration between the bodies, assume that the various bodies move by a distance x1, x2, ...and so on.Calculate the number of segments in the rope.
Step 4: Relate the distance moved. (Total change in the length of the rope must be zero.) To do these calculate the change in length of each segment of the rope. Then add these changes to get the total change in the rope.
Step 5: Once we get the relation between the distances moved, the acceleration relation will be the same

 

[jmb]

How are these two pulleys connected? I don't think the string throught pulley 2 connects to m1 or else the tension and acceleration would be the same. I am having a hard time visualizing what is happening.

As I understand the diagram, the string through pulley 2 connects to the center of pulley 1. That is, the extension of m2 controls the position of pulley 1. The two strings are not connected directly.

Sorry if this is stating the obvious (!), but I thought from your wording it might be what was confusing you...

 

[blackboy]

As I understand the diagram, the string through pulley 2 connects to the center of pulley 1. That is, the extension of m2 controls the position of pulley 1. The two strings are not connected directly.

Sorry if this is stating the obvious (!), but I thought from your wording it might be what was confusing you...

You weren't stating the obvious, it was just what I needed to know. Ok so when m2 moves, it causes P1 to move. The net force acting on m2 is m2g-T2(Where T2 is the tension in that string). So m2g-T2=m2a2. T2=m2(g-a2). Let's say the tension in the string of P1 is T1. Because it is accelerating T2-T1=P1a=m1a. Simplifying I get T1=m2(g-a2)-m1a1. I try to get everything in terms of g m1 m2, but I keep going in circles.

@rl.bhat- Can you show me what you mean? I don't really understand your post.

 

[rl.bhat]

@rl.bhat- Can you show me what you mean? I don't really understand your post.

In the problems there are two ropes, two pulleys and two masses.
The rope1, which passes through P₁, fixed at one end and other end is free to move with m₁
The rope2, which passes through P₂, connected to m₂ at one end and other end is connected to P₁.
When m₂ moves down due to gravity, segment 1 (P2m2) increases by x1 and segment 2 (P1P2) decreases by x2.
Net change in the length of rope 1 = x1 - x2 = 0 or x1 = x2. Hence a1 = a2.
The tension in each segment = T
When the P1 moves towards right, segment 3, one end of which is fixed to the wall, increases by x2 and the segment 4, which is connected to m1, decreases by x2 - x3, because m1 is moving in the same direction as P1.
Hence net change in rope 2 = x2 + x2 - x3 = 0 or 2x2 = x3 or 2a2 = 2a1 = a3.
Tension in each segment = T/2.
Now you can solve the rest of the problem.

 

[jmb]

I just wanted to elaborate on rl.bhat's answer...

Tension in each segment = T/2.

In case it isn't clear, this last statement is due to force balance on the pulley P1 (it would be different if the pulley had non-zero mass).

In fact another way to deal with these problems is to write down the force balance for every movable object (masses and pulleys) and then use inextensibility of the ropes.

Let us know how you get on with the last part of the problem, or if you have trouble understanding anything already said.

 

[blackboy]

Well the thing I don't understand it how can you make the assumption that segment 2 decreases by x2 and segment 1 increases by x1. So if m2 moves 2m, then segment 1m? How did you get this? I am sorry I don't understand.

 

[rl.bhat]

Well the thing I don't understand it how can you make the assumption that segment 2 decreases by x2 and segment 1 increases by x1. So if m2 moves 2m, then segment 1m? How did you get this? I am sorry I don't understand.

When m₂ is not moving
the distance between m₂ and P₂ = d₁
the distance between P₂ and P₁ = d₂
the distance between m₁ and P₁ = d₃
the distance between P₁ and wall = d₄
When the mass m₂ moves down through x₁
d₁ becomes d₁ + x₁...> change x₁.
d₂ becomes d₂ - x₂...> change -x₂
So net change = x₁ - x₂ = 0. Or x₁ = x₂
d₃ becomes d₃ + x₂ -x₃...> change x₂- x₃
d₄ becomes d₄+ x₂...>change ...x₂
Net change 2*x₂ -x₃ = 0 or 2x_{2= x3}

 

[jmb]

If you want to understand it physically rather than mathematically (although really you should make sure you are happy with both explanations) you can think of it like this...

Questions:

• What does moving m₂ do to P₁?
• What does moving P₁ do to the length of rope between P₁ and the wall?
• What does this do to the length of rope between P₁ and m₁?
• So what is the net effect on m₁?

(Try and think through these yourself before checking the answers below...)

Answers:

• m₂ is directly attached to P₁ by an inextensible rope, so if m₂ moves by a distance $$\Delta x$$ then P₁ also moves by a distance $$\Delta x$$.
  
• Since P₁ moves by $$\Delta x$$ the length of rope between the wall and P₁ also increases by $$\Delta x$$.
  
• This rope is also inextensible so it means that length of rope between m₁ and P₁ (part of the same piece of rope) must decrease by $$\Delta x$$, bringing m₁ closer to P₁ by a distance $$\Delta x$$.
  
• Since m₁ is now closer to P₁ by $$\Delta x$$ and P₁ is itself $$\Delta x$$ further away from the wall, then m₁ must have moved a total distance $$\Delta x + \Delta x = 2\Delta x$$ away from the wall.