Tension problem: 2 ropes, 3 blocks

[JoshBuntu]

Homework Statement

The diagram is here http://img375.imageshack.us/img375/5930/tensionproblem.png

What is the tension in the cord connecting B and C? How far does A move in the first 0.250 seconds of movement assuming it does not reach the pulley?

Homework Equations

F=ma

The Attempt at a Solution

Tension1=(Mass of A)(acceleration)
(Mass of B)(gravity) - Tension1=(Mass of B)(acceleration)
I assumed that the tension in each equation was equal and so set the equations equal, solving for (acceleration):
(acceleration) = [(mass of B)(gravity)]/[(mass of A)(mass of B)] = 3.8 m/s^2

And honestly I just don't know what the hell I'm doing. I've been at this for over 3 hours. I tried to teach this tension thing to myself. What's also really confusing me is that cord between B and C! I don't understand any of the concepts and I don't have anyone to help me, and I also don't know whether or not I'm on the right track. Please help ):

 

[PhanthomJay]

And honestly I just don't know what the hell I'm doing.

I like your honesty

I've been at this for over 3 hours. I tried to teach this tension thing to myself. What's also really confusing me is that cord between B and C! I don't understand any of the concepts and I don't have anyone to help me, and I also don't know whether or not I'm on the right track. Please help ):

Its all about drawing good free body diagrams of each block, identifying the forces acting on each block, including tension and weight forces, and then applying Newton's laws. Hint: Tension forces always pull away from the body on which they act. The tension in a cord wrapped around an ideal pulley is the same on both sides of that pulley. They are not necessarily the same when connected to each side of a block (2 different cords). The magnitude of the accelerations of all blocks are the same. Are you familiar at all with free body diagrams?

 

[ehild]

Tension1=(Mass of A)(acceleration)
(Mass of B)(gravity) - Tension1=(Mass of B)(acceleration)

Your first equation is correct, but the tension of the second rope between B an C is missing from the second equation. The tension of the second rope is different from the tension in the first rope. And you have a third equation for mass C.

When you have all three equations, you can add them together and all tensions will cancel and you get the acceleration. From the acceleration, you can calculate the tensions.


ehild

 

[jhae2.718]

You should look at the system in three parts, blocks A-C. What forces act on each?

You'll have two tension forces because there are two different ropes. For your equation for block B, you need to add the second tension force. You need a third equation for block C.

Edit: way too late...

 

[JoshBuntu]

Ok, so, for free body diagrams:

Mass A, normal force acting up and weight acting down which cancel each other, and Tension1 acting to the right

Mass B, Weight acting down and Tension1 acting up, weight greater than Tension1(?)

Mass C, Weight acting down and Tension1 acting up, weight greater than Tension2(?)

ehild: What do you mean? Should I treat blocks B and C as one mass (adding their masses and treating as a single block)? Also, would like, gluing blocks B and C together be different than having them attached by a rope?

 

[jhae2.718]

On B, you also have the second tensile force acting downwards. Mass C has tension 2 acting upwards. (T1 =\= T2!)

Once you have your three equations, you can solve them like any system of equations to find the acceleration, and then the tensions.

 

[JoshBuntu]

:O Ok I will try that right now! Don't leave me!

 

[JoshBuntu]

Ok so tell me if I'm right!

The second tension (Tension2) between B and C is (Tension2)=(mass of C)(acceleration)

and (Tension1)=(mass of A)(acceleration)

so I think the correct form of the second equation that I previously got wrong is:
(mass of B)(gravity)+(Tension2) - (Tension1) = (mass of B + mass of C)(acceleration)

and substituting for Tension2:
(mass of B)(gravity)+[(mass of C)(gravity)-(mass of C)(acceleration)] - (Tension1) = (mass of B + mass of C)(acceleration)

?

 

[jhae2.718]

For the second tension, you need to take the weight of the block into account.

 

[JoshBuntu]

THERE is where I have a problem! block C is hanging on the rope, so it's creating a tension. So like, block B is kind of acting like a ceiling that block C is hanging from by a rope - how does block B constitute a tension to the second tension? Isn't block B just falling while block C is pulling the rope down? ):

 

[jhae2.718]

All of the ropes are in tension, so if we "cut" the ropes, there will be a tension force pulling on each block. Note that the blocks are all moving with the same speed.

 

[ehild]

O

ehild: What do you mean? Should I treat blocks B and C as one mass (adding their masses and treating as a single block)? Also, would like, gluing blocks B and C together be different than having them attached by a rope?

You can imagine that B and C are glued together, but then your second equation must include the sum of masses of B and C.

If you read the text of the problem it asks the tension in the rope connecting B and C, So it is better to include it and handle the three masses separately.

If you draw a proper free body diagram, you will see the forces acting on B are:
M_Bg +T₂-T₁=M_Ba

and those for C are the weight of C (down) and T₂ (up)

ehild

 

[JoshBuntu]

Woah, woah, I think I got this. Is the final formula for Tension2:

Tension2 = [(mass of A)(mass of C)(gravity)]/[(mass of C) + (mass of B) + (mass of A)]

with the answer for tension 2 being 36.8 Newtons?

 

[JoshBuntu]

And thank you ehild! Now I get it, I think. Tension2 is pulling down on block B along with the weight of block B. These 2 forces together are greater than Tension1. okokokok. Before I wasn't really getting what tension 2 was actually doing it. It's pulling down on block B and pulling up on block C with equal magnitudes, right??

 

[ehild]

Woah, woah, I think I got this. Is the final formula for Tension2:

Tension2 = [(mass of A)(mass of C)(gravity)]/[(mass of C) + (mass of B) + (mass of A)]

with the answer for tension 2 being 36.8 Newtons?

You have a typo, there must be a "+" between mass of A and mass of B in the numerator. The 36.8 N for tension 2 is OK.

And yes, a massless rope has the same tension along all this length, pulling the bodies at both ends with forces of equal magnitude and opposite direction along the rope.

ehild

 

[JoshBuntu]

Yay! (: But WAIT? "there must be a "+" between mass of A and mass of B in the numerator" There is no mass of B in the numerator... was I supposed to have one there? Because I solved the equation just as I typed it here with only multiplication in the numerator and got 36.8 Newtons...

 

[ehild]

You are right, it is the right formula, right result, I was not quite awake when I replied

ehild