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Tension between two accelerating blocks

  • Thread starter Fizic
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  • #1
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Problem text:

Forces act on block A and block B, which are connected by a massless string. Force [itex]\vec{F}[/itex]A=(12 N)[itex]\hat{i}[/itex] acts on block A, with mass 4.0 kg. Force [itex]\vec{F}[/itex]B=(24 N)[itex]\hat{i}[/itex] acts on block B, with mass 6.0 kg. What is the tension in the string?

The system looks something like this:

[A]-----

Relevant equations:

F=ma

Attempt at solution:

The professor never went over tension in class, so I've been Googling the pants off this problem and combing through the book for a solution, each to no avail. I tried working through it for about 30 minutes and gave up in frustration, after having come up with about 6 different answers.
 

Answers and Replies

  • #2
Doc Al
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Show what you've tried.

Hint: Identify the forces acting on each mass and apply Newton's 2nd law.
 
  • #3
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Posting anything I've done wouldn't be of value, since everything I tried was essentially the mathematical equivalent of alchemy.

If by apply Newton's 2nd law, you mean add up all the forces, then I come up with Fnet=36N, M=10kg, and A=3.6m/s^2.
 
  • #4
Doc Al
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If by apply Newton's 2nd law, you mean add up all the forces, then I come up with Fnet=36N, M=10kg, and A=3.6m/s^2.
That's fine. You took as your 'system' both masses together. But to solve for the tension, now analyze the forces on one of the blocks.
 
  • #5
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That's fine. You took as your 'system' both masses together. But to solve for the tension, now analyze the forces on one of the blocks.
So...

T=3.6m/s^2*14.4kg=14.4N

Is it really that simple? I'm gonna have to kick myself for that!

Anyways, thanks a lot Doc Al!
 
Last edited:
  • #6
Doc Al
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So...

T=3.6m/s^2*14.4kg=14.4N
Not sure what you are doing here. (Does that equation even make sense? :wink:)

Do this:
Pick one of the masses.
Identify the forces acting.
Apply Newton's 2nd law: ƩF = ma. (You already found the a!)
 
  • #7
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Not sure what you are doing here. (Does that equation even make sense? :wink:)

Do this:
Pick one of the masses.
Identify the forces acting.
Apply Newton's 2nd law: ƩF = ma. (You already found the a!)

Woops. Meant to write T=3.6m/s^2*4kg=14.4N

That should be right, no?
 
  • #8
Doc Al
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Woops. Meant to write T=3.6m/s^2*4kg=14.4N

That should be right, no?
What you have is m*a for the 4kg mass. That's the right hand side of the equation ƩF = ma. But what about the left hand side?

You need to identify the individual forces acting on that mass (one force will be the tension) and then add them to get an expression for ƩF.
 
  • #9
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What you have is m*a for the 4kg mass. That's the right hand side of the equation ƩF = ma. But what about the left hand side?

You need to identify the individual forces acting on that mass (one force will be the tension) and then add them to get an expression for ƩF.
So you're saying I need to take into account the 12 N and subtract it from the 14.4 N I just calculated to get 2.2 N of tension force?

Edit:

Nevermind, that was more alchemy. My give up
 
  • #10
Doc Al
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Do this: Name the two forces acting on the 4kg mass and tell me the direction they act.
 
  • #11
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Do this: Name the two forces acting on the 4kg mass and tell me the direction they act.
What I know for sure is that there is a 12N force in the positive direction, and I'm fairly sure that a 14.4N force is being applied, but in which direction I don't know.

Edit:

Here's another try.

[itex]\vec{F}[/itex]T + [itex]\vec{F}[/itex]A=ma

ma-[itex]\vec{F}[/itex]A=[itex]\vec{F}[/itex]T

4kg*3.6m/s2-12N=[itex]\vec{F}[/itex]T

14.4N-12N=[itex]\vec{F}[/itex]T=2.4N
 
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  • #12
Doc Al
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What I know for sure is that there is a 12N force in the positive direction,
Good. That force is given.

and I'm fairly sure that a 14.4N force is being applied, but in which direction I don't know.
Don't be guessing the answer--you will solve for it. Just label the force and its direction. (What's exerting the force?)
 
  • #13
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Good. That force is given.


Don't be guessing the answer--you will solve for it. Just label the force and its direction. (What's exerting the force?)
[itex]\vec{F}[/itex]i=(2.4N)[itex]\hat{i}[/itex]

Final answer.
 
  • #14
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Again, I don't want the final answer. I want to see you solve for the final answer. (No guessing allowed!)

Hint: The string exerts a tension force--call it T--on the mass. In what direction does the tension pull on the 4 kg mass?

Then write an expression for ƩF, the sum of the forces on that mass.
Wait a sec... Isn't that what I did in post #11? Maybe you didn't see; I edited it.
 
  • #15
Doc Al
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Here's another try.

[itex]\vec{F}[/itex]T + [itex]\vec{F}[/itex]A=ma

ma-[itex]\vec{F}[/itex]A=[itex]\vec{F}[/itex]T

4kg*3.6m/s2-12N=[itex]\vec{F}[/itex]T

14.4N-12N=[itex]\vec{F}[/itex]T=2.4N
Now you're cooking! :approve:
 
  • #16
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Thanks to you. srsly
 
Last edited:

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