Pulleys, tension, and blocks oh my

[grepory]

Homework Statement

In the figure, the pulleys and the cords are light, all surfaces are frictionless, and the cords do not stretch.
(a) How does the acceleration of block 1 compare with the acceleration of block 2? (Use a_2 for a2.)
(b) The mass of block 2 is 1.22 kg. Find its acceleration as it depends on the mass
m1 of block 1. (Use m_1 for m1.)
(c) Evaluate your answer for m1 = 0.537 kg. Suggestion: You may find it easier to do part (c) before part (b).
(d) What does the result of part (b) predict if m1 is very much less than 1.22 kg?
(e) What does the result of part (b) predict if m1 approaches infinity?
(f) What is the tension in the long cord in this last case?
(g) Could you anticipate the answers (d), (e), and (f) without first doing part (b)? Explain your answer.

Homework Equations

F = ma

The Attempt at a Solution

Alright.. so here's my force diagram.

http://binary-snobbery.com/pics/physics09072009.jpg

I labelled all of the tensions the same, because... I don't know. I just figured they'd all be the same... with the exception of the 2T which is just T+T.

A/B) a₁ = 2a₂. I don't really know why, but it was my best guess, because m2 has 2T acting as an opposing force to its natural tendency toward motion.

C) I tried this:
m1:
$$\Sigma$$ F_x = T = m₁a₁

m2:
$$\Sigma$$ F_y = 2T - m₂g = m₂a₂

So then I used the identity found in part A to solve the system for a₂.

2(m₁a₁) - m₂g = m₂a₂
2(m₁(2a₂)) - m₂g = m₂a₂
4m₁a₂ - m₂g = m₂a₂

Do a little algebra and you end up with:
a₂ = m₂g / 4(m₁ - m₂)

I plugged in the given values, but I got the wrong answer. That's basically where I gave up. I am missing something fundamental, I guess.

 

[rl.bhat]

For m2 it should be
Fy = m2*g - 2T.

 

[grepory]

For m2 it should be
Fy = m2*g - 2T.

Why? That's completely counterintuitive.

If you think of going UP as positive acceleration on the y-axis and going DOWN as negative acceleration (as I think most of the physics world does), then g is -9.8m/s² and not 9.8m/s². Hence,

F_y = 2*T - m₂g

 

[rl.bhat]

Why? That's completely counterintuitive.

If you think of going UP as positive acceleration on the y-axis and going DOWN as negative acceleration (as I think most of the physics world does), then g is -9.8m/s² and not 9.8m/s². Hence,

F_y = 2*T - m₂g

In that case Fy should be negative, because it is in the direction of g

 

[grepory]

In that case Fy should be negative, because it is in the direction of g

Well, I just thought about how "Direction" in the case of acceleration is whether or not the velocity is decreasing or increasing... Not so much the "direction" of the acceleration. I guess you could say that if something is slowing down, it's accelerating in the opposite direction that it's headed. Anyway.. I think really we're splitting hairs.

I guess what I'm trying to say is that... I'm not sure what you're trying to say.

 

[rl.bhat]

When the system is released, in which direction m2 will move?
If it moves in the down ward direction what will be its acceleration?
When you drop a body, what happens to it velocity? Increases or decreases?
If it increases what is the sign of g?

 

[ideasrule]

C) I tried this:
m1:
$$\Sigma$$ F_x = T = m₁a₁

m2:
$$\Sigma$$ F_y = 2T - m₂g = m₂a₂

The problem is that Fy is a negative number if you define it as 2T-m2g instead of m2g-2T. That means a2 would be negative, which means a1 doesn't equal 2*a2; it's equal to -2*a2.

I usually like to keep all forces positive; it's much more intuitive that way.

 

[OSearcy4]

all this discussion can someone answer part A?

 

[ideasrule]

The original poster already gave the right answer for part A:

a1=2a2

He made a mistake with the signs later on, but there's no problem with that equation.

 

[grepory]

The problem is that Fy is a negative number if you define it as 2T-m2g instead of m2g-2T. That means a2 would be negative, which means a1 doesn't equal 2*a2; it's equal to -2*a2.

I usually like to keep all forces positive; it's much more intuitive that way.

But forces are vectors :(

I looked at the problem some more tonight, and I'm just lost. I setup an appointment with a tutor tomorrow (fortunately, our department has free personal tutoring). I'm hoping they can help shed some light on it.

 

[OSearcy4]

yea i read it after i posted my bad
but how do you figure part b

 

[OSearcy4]

what school do you go to

 

[grepory]

yea i read it after i posted my bad
but how do you figure part b

Well.. If I'm reading the responses, so far, correctly... I was on the right track. If you know that a₁ = 2*a₂, then you can use that identity to get an equation out of the force equations for each of the masses. I still don't understand where that identity comes from, though.

 

[OSearcy4]

i still can't figure out b.
if anyone can help please email me OSearcy4@aol.com

 

[rl.bhat]

i still can't figure out b.
if anyone can help please email me OSearcy4@aol.com

The acceleration in m₂ is given by
a₂ = (m2*g - 2T)/m₂
But T also equal to m₁*a₁
Substitute this value in the above equation and put a₁ = 2*a₂, find the value a₂.

 

[grepory]

The acceleration in m₂ is given by
a₂ = (m2*g - 2T)/m₂
But T also equal to m₁*a₁
Substitute this value in the above equation and put a₁ = 2*a₂, find the value a₂.

After thinking about it for a while today, I figured out part a..

The distance that m2 travels downward as m1 travels to the right is 1/2 the distance traveled by m1, i.e.
x_1 = 1/2 x_2
because of the pulley situated above m2 and its connection to a stationary object on the other side of the gap.

They each move this distance (x, and .5x respectively in the same amount of time). This means that their velocities would be of the form:
v1 = xt and v2 = x/2 * t

If you take the derivative of these velocity functions wrt to time, you arrive at:
a1 = x and a2 = x/2

Solve a2's equation for acceleration for x, and you are presented with the relationship between a1 and a2:
a1 = 2a2.

What I'm stuck on is the tension in the rope...

We're assuming that the tensions on either side of the lower pulley (above m₂) are equal to each other. Why can we make that assumption?

 

[grepory]

We're assuming that the tensions on either side of the lower pulley (above m₂) are equal to each other. Why can we make that assumption?

Also, I've finished the problem... but I don't understand why the tensions are equal.

 

[ideasrule]

Also, I've finished the problem... but I don't understand why the tensions are equal.

It's because the rope is assumed to be massless. Consider the first cm of the rope; because ma=0, the tension pulling it forwards must be equal to the tension pulling it backwards. The same argument applies to the 2nd cm, and the 3rd, and so on. It follows that tension is the same throughout a massless rope. Real ropes are amazingly light for their strength, so this approximation works well in the real world.

 

[grepory]

It's because the rope is assumed to be massless. Consider the first cm of the rope; because ma=0, the tension pulling it forwards must be equal to the tension pulling it backwards. The same argument applies to the 2nd cm, and the 3rd, and so on. It follows that tension is the same throughout a massless rope. Real ropes are amazingly light for their strength, so this approximation works well in the real world.

Oh. I'd sort of intuited that, but I was pretty doubtful of myself. I was imagining the two lengths of the rope to the right and left of the lower pulley, but I couldn't decide what I thought about the tension. Thinking about it again, though... it makes perfect sense.

Thanks for all the help!

 

[MechaMZ]

After thinking about it for a while today, I figured out part a..


They each move this distance (x, and .5x respectively in the same amount of time). This means that their velocities would be of the form:
v1 = xt and v2 = x/2 * t

I think it should be :

displacement of m₁ = x
displacement of m₂ = 0.5x

then,

V₁ = x₁ / t => xt^-1
dv/dt = -1xt^-2
a₁ = -xt^-2

V₂ = 0.5x / t => 0.5xt^-1
dv/dt = -0.5xt^-2
a₂ = -0.5xt^-2

then a₁ / a₂ = 2/1
then a₁ = 2a₂

please correct me if I am wrong.

by the way, I also have no idea how to figure out the displacement of m₁ is equal to half of m₂. Hope someone could kindly explain to me =)

 

[grepory]

I think it should be :

displacement of m₁ = x
displacement of m₂ = 0.5x

Yeah.. That is a much more rigorous approach to it!

by the way, I also have no idea how to figure out the displacement of m₁ is equal to half of m₂. Hope someone could kindly explain to me =)

That's one of the easier parts. Think about the displacement of m₁ being in terms of the length of rope that has moved over into m₂'s subsystem with its pulley. If you add "X" meters of rope to that system, how much does m₂ actually move? It may also make it easier if you forget about the length of rope to the right of m₂'s pulley that's _above_ the rope coming from m₁ if that makes sense.

 

[MechaMZ]

Thank you for your explanation. but for the question below, how you assume the tension in the system? why there is only one tension force?

the man and the platform together weigh 960 N. The pulley can be modeled as frictionless. Determine how hard the man has to pull on the rope to lift himself steadily upward above the ground.

http://img141.imageshack.us/img141/236/p433.gif

I thought there should be 2 tension forces, since the man is pulling the other end of the rope? Isn't because the man is considered as part of the system with the platform and pulley?

http://img269.imageshack.us/img269/70/pulley.jpg

 

[grepory]

I thought there should be 2 tension forces, since the man is pulling the other end of the rope? Isn't because the man is considered as part of the system with the platform and pulley?

There is tension on the right-hand side, and on the left-hand side, there is the force of the man pulling on the rope. And yes, I think it has something to do with the fact that the man is standing on the platform with the pulley.

 

[MechaMZ]

Hi grepory,

i just found a proper method to prove the ratio of a1 and a2, it could be done with work done involved. i show you the method tomorrow