3 pulleys - 2 masses on incline plane

[Saints-94]

Homework Statement

http://tinypic.com/r/qs9q4o/9 http://tinypic.com/r/qs9q4o/9
Verify that the log will move up the ramp under the given conditions

Homework Equations

The Attempt at a Solution

m1 = 220

m2 = 130

mu = 0.4

theta = 25

g = 9.81

N = (m1*g)*cos(theta)1) m2g – T = m2a Forces acting on Block B

2) 2T – m1g*sin(theta) – mu*N = m1a Forces acting on log1) T = m2g – m2a

2) 2 (m2g – m2a) – (m1g*sin(theta)) – mu*N = m1a2m2g – 2m2a – (m1g*sin25) – (0.4 * N ) = m1a

(220*130*9.81) – (220*9.81*sin25) – (0.4*1956) = m1a + 2m2a

(2550.6) – (912.1) – (782.4) = a(m1+2m2)

856.1 / (m1+2m2) = a

856.1 / (220+260) = aa = 1.78m/s^2
Unsure about the forces acting on the double pulley. I am right to assume it is 2T? Does this give me the correct value of 'a' to assume the log is accelerating up the slope? [/B]

 

[Saints-94]

Yes. Is that incorrect?

 

[haruspex]

Yes. Is that incorrect?

At the same time as you replied, I deleted my post because I noticed you later had 2a terms, so I thought I should check whether you had defined a as the acceleration of one and were using 2a for the acceleration of the other. But having checked, no, it seems you did use a for both.
The rope is constant length. As the weight descends a distance y, what happens to the three sections of rope?

 

[Saints-94]

At the same time as you replied, I deleted my post because I noticed you later had 2a terms, so I thought I should check whether you had defined a as the acceleration of one and were using 2a for the acceleration of the other. But having checked, no, it seems you did use a for both.
The rope is constant length. As the weight descends a distance y, what happens to the three sections of rope?

The rope carrying Block B increases as the block lowers. The two lengths of rope carrying the log stay equal to each other and shorten. Is that what you were asking?

 

[haruspex]

The rope carrying Block B increases as the block lowers. The two lengths of rope carrying the log stay equal to each other and shorten. Is that what you were asking?

Yes, but a bit more detail... if block B descends a distance y, how much do the two lengths holding the log shorten by?
What does that tell you about accelerations?

 

[Saints-94]

If block B decends by 1m, then the log will move 2m due to the two pulleys?

 

[haruspex]

If block B decends by 1m, then the log will move 2m due to the two pulleys?

No. One step at a time, no wild guesses please.
You wrote, correctly, that the two ropes parallel to the slope remain the same length as each other. We also know the total length of the rope is constant. Write that as an equation relating the lengths of the sections.

 

[Saints-94]

Is the rope pulling the log two times the length of the rope supporting the block?

 

[Carson Birth]

I got nearly the same the answer (rounding error) so that could be a good sign. I treated the force on the double pulley as 2x like you, I am pretty sure that's the way to do it, I could be wrong though. A little tip, I like to break down each force into each direction, form a sum of forces in each direction then form a triangle of the resultant force (being the hypotenuse), just the way my teacher taught me. It makes it nice and simple for figuring out which force is going where.

 

[haruspex]

Is the rope pulling the log two times the length of the rope supporting the block?

I said no wild guesses.
Let the total length of the rope be L.
If the vertical length is y, how long are the other two sections?

 

[Saints-94]

I believe the other two sections would both be y also?

 

[Carson Birth]

Is the rope pulling the log two times the length of the rope supporting the block?

If we are saying that force from the block is doubled due to the set up of the double pulley then something has to compensate... the distance moved.

 

[haruspex]

I believe the other two sections would both be y also?

That would make the total length 3y, but, since the total length is constant, y would be constant, so the weight cannot move?
Please, please, please try writing equations to represent the clear facts:
The total length of the string is constant;
The two lengths parallel to the slope are equal in length.
I will not respond further until you show an attempt to do that.

 

[Saints-94]

If we are saying that force from the block is doubled due to the set up of the double pulley then something has to compensate... the distance moved.

Yes i agree. Would this affect the acceleration? Would it make the acceleration 2*a at the log?

 

[Saints-94]

That would make the total length 3y, but, since the total length is constant, y would be constant, so the weight cannot move?
Please, please, please try writing equations to represent the clear facts:
The total length of the string is constant;
The two lengths parallel to the slope are equal in length.
I will not respond further until you show an attempt to do that.

Previously you stated the total length of the rope was L and vertical was y. In my equations i have labelled the forces acting on the ropes as 2T at the log and T acting against the block. I don't understand what you're asking?

 

[haruspex]

Previously you stated the total length of the rope was L and vertical was y. In my equations i have labelled the forces acting on the ropes as 2T at the log and T acting against the block. I don't understand what you're asking?

It is not (directly) related to the forces. It is just a matter of the way the lengths of the sections relate to each other.
If each of the sections parallel to the slope is length x, and the vertical section is length y, what equation relates x, y and the total length L?

 

[Saints-94]

It is not (directly) related to the forces. It is just a matter of the way the lengths of the sections relate to each other.
If each of the sections parallel to the slope is length x, and the vertical section is length y, what equation relates x, y and the total length L?

Is it as simple as 2x+y=L?

 

[Carson Birth]

Yes i agree. Would this affect the acceleration? Would it make the acceleration 2*a at the log?

Think about the equation of acceleration, distance/time squared, if the log is moving less distance in the same time then...

 

[Saints-94]

Think about the equation of acceleration, distance/time squared, if the log is moving less distance in the same time then...

It is accelerating slower?

 

[haruspex]

Is it as simple as 2x+y=L?

Yes!
Now, what does that tell you about the accelerations? What do you do to a distance variable to turn it into an acceleration variable?

 

[Saints-94]

Yes!
Now, what does that tell you about the accelerations? What do you do to a distance variable to turn it into an acceleration variable?

I'm not sure, I haven't come across this before.

 

[haruspex]

I'm not sure, I haven't come across this before.

Have you been taught any differential calculus? Like, velocity = rate of change of position?

 

[Saints-94]

Have you been taught any differential calculus? Like, velocity = rate of change of position?

V^2 = U^2 + 2as?

 

[haruspex]

V^2 = U^2 + 2as?

No.
Does v =dx/dt ring any bells?

 

[Saints-94]

No.
Does v =dx/dt ring any bells?

Yes. But I'm unsure how to apply it to find acceleration.

 

[haruspex]

Yes. But I'm unsure how to apply it to find acceleration.

Then let's try to reason it out without calculus.
You found that 2x+y=L. So if the weight descends 1m, how far does the log move, and in which direction?

 

[Saints-94]

Then let's try to reason it out without calculus.
You found that 2x+y=L. So if the weight descends 1m, how far does the log move, and in which direction?

It will move up the slope. I'm unsure what distance, but would it be 2m?

 

[haruspex]

It will move up the slope. I'm unsure what distance, but would it be 2m?

Try it in the equation. We've increased y by 1. You are suggesting x will decrease by 2, yes? Does that still give a total of L?

 

[Saints-94]

Try it in the equation. We've increased y by 1. You are suggesting x will decrease by 2, yes? Does that still give a total of L?

(y+1) + (2x-2) = L ?

 

[haruspex]

(y+1) + (2x-2) = L ?

No, you suggested the log would ascend 2m, which means each x would lose 2m. That would mean you start with L=y+2x, And end with a string length (y+1)+2(x-2)=y+2x-3=L-3. The string has mysteriously shrunk by 3m.
Try another way of changing x when y increases by 1m.