# Homework Help: Max acceleration w/ a tension limit

1. Sep 27, 2008

### j123

1. The problem statement, all variables and given/known data

A 710kg boulder is raised from a quarry 191m deep by a long uniform chain having a mass of 595kg. This chain is of uniform strength, but at any point it can support a maximum tension no greater than 2.80 times its weight without breaking.

How do i find the maximum acceleration in order to get it out of the quarry?

Attempted but all wrong... just need to be pointed in the right direction. the 2.8x is confusing. and im lost to what equations i should use now.

know the Force includ gravity for 710kg is 6958N, and 595kg is 5831N...

Last edited: Sep 27, 2008
2. Sep 27, 2008

### Staff: Mentor

What's the weight of the chain? What's 2.8 times that weight? That's the maximum allowable tension.

Hint: Where along the chain will the tension be greatest?
Apply Newton's 2nd law.

3. Sep 27, 2008

### j123

-the weight of the chain is w=mg so 595kg*9.8= 5831N x 2.8 times is = 16326.8N.?

-where the boulder and the chain connects, i would say.

-Newton's 2nd law: EFx=ma.x EFy=ma.y (how many steps do i apply? and how?, I think only EFy is only used since it's going up) so Fy/m = a.

Last edited: Sep 27, 2008
4. Sep 27, 2008

### Galileo's Ghost

Remember, with Newton's Second Law, the EFy you have written represents the SUM of forces in the y direction. Identify all of those forces and include them in the equation!

Also, rethink the "where the boulder and the chain connects" statement!

5. Sep 27, 2008

### j123

-so EFy is 6958N of boulder + 5831N of chain =12789N (isn't this divided by the two masses equal back to g?)

-the max tension is 16326.8N for the chain

-isn't the tension the greatest where the boulder pulls on the chain?

6. Sep 27, 2008

### Galileo's Ghost

Don't forget to include tension in your sum! Also, pay attention to the direction of each of these forces as you set up the Second Law equation.

As far as the greatest tension, doesn't the boulder really just pull on a single link at the end of the chain? But then doesn't that link pull on the next link and that one pulls on the next link and so on... How would the magnitude of all these link pulls compare?

7. Sep 27, 2008

### j123

-OK, (i thought the tension was these two forces) im actually really fustrated and completely out of this. Can you explain? and am i even close to the end?

i kinda gave up on this problem already, i really just want to know what im doing wrong.

(will check on this later today)

Last edited: Sep 27, 2008
8. Sep 27, 2008

### Staff: Mentor

Good.
Think about it. The bottom link of the chain has to support the weight of the boulder, but the next highest link has to support not just the weight of the boulder but the weight of the chain beneath it as well.

Hint: Apply Newton's 2nd law to the entire system of "chain + boulder".

9. Sep 27, 2008

### j123

-so the greatest tension is at the top of the chain.

-so EFy is 6958N of boulder + 5831N of chain =12789N, right? (someone else said i should include the tension but i thought these were the only two forces and they are the T also)

10. Sep 27, 2008

### Staff: Mentor

Right.
Those are the weights of the boulder and chain, which act downward. Don't forget the force pulling up on the chain, a force equal to the tension at the top of the chain.

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