# Chain with 2 links lifted vertically - Force

1. May 8, 2010

### mybrohshi5

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

A chain consisting of two links, each of mass 0.5 kg, is lifted vertically with an acceleration of 3.0 m/s2 upward.

The magnitude of the downward force exerted on the top link by the bottom link is?

2. Relevant equations

F = ma

3. The attempt at a solution

F1,2 = The force on 1 by 2

The Fnet on either of the two links is (0.5kg)(3m/s2) = 1.5N

so

Fnet = Fpull - Fg - F1,2

Fpull = 1kg(3m/s2) = 3N

so

1.5N = 3N - (.5kg)(9.8m/s2) - F1,2

F1,2 = 3.4 N downward

The answer is 6.4N but i dont understand why? where did i go wrong?

this is exactly what i did on a homework assignment and got it right but this is what i did on the test and i got it wrong since the answer is 6.4N not 3.4N like i got.

Thanks for any help

2. May 8, 2010

### PhanthomJay

You didn't calculate F_pull correctly. Where's the weight force? You might want to check your new result by drawing a FBD of the bottom link.

3. May 8, 2010

### mybrohshi5

I dont see how F_pull is wrong?

The only other thing i can think of doing to calculate this is to factor in the the acceleration of gravity?

and by weight force do you mean Fg cause i have that in there....

4. May 8, 2010

### mybrohshi5

What i am really confused about is i used this same method on a mastering physics problem and got all of the answers right so i guess i got lucky on that. I really need to figure out how to do this right because im sure something similar to it is going to be on my final.

5. May 8, 2010

### PhanthomJay

You have got to draw free body diagrams to solve these problems. You noted that
But F-pull is the pulling force which must be greater than the weight of the links in order to accelerate them upward at 3m/s/s. Since the links weigh about 9.8 N, the pull force can't be just 3 N.
Again, you can also calculate F_1,2 directy by drawing a FBD of the botton link alone.

6. May 8, 2010

### mybrohshi5

oh that makes sense.

ok so using my fbd of the botton link i have the force pulling it up which i believe is (3+9.8)(1) = 12.8 N in the upward direction

there is also the F_2,1 force in the upward direction (this is equal to F_1,2)

then there is the force of gravity of link 2 going downwards.

Does that seem right?

7. May 8, 2010

### PhanthomJay

No. If you draw a FBD of the bottom link, there are only two forces acting on it, its weight down, and the F-2,1 force acting up. Use newton 2 to solve for F_2,1. Don't put in forces that don't exist when you draw your free body diagram.

8. May 8, 2010

### mybrohshi5

Oh jeez i feel stupid. That was a lot easier then i was making it.

so sum of the forces in the y = ma

Fy = 0.5kg(9.8+3)

Fy = 6.4 N

Thank you Jay