# Newton's 2nd Law

1. Sep 9, 2009

### Zhalfirin88

1. The problem statement, all variables and given/known data
A chain consisting of five links, each of mass 0.090 kg, is lifted vertically with a constant acceleration of a = 3.6 m/s2. Find the magnitude of the force that link 3 exerts on link 2.

Note: The chains are connected vertically

2. Relevant equations
$$\Sigma F = ma$$

3. The attempt at a solution
How would I calculate the force? I've tried the force of gravity minus the force of 3 chain links but that didn't work. Numerically I did: I called down the positive direction.

$$(3 * .09kg)(9.8 m/s^2) - (3 * .09kg)(3.6 m/s^2)$$

Last edited: Sep 9, 2009
2. Sep 9, 2009

### Fightfish

$$\Sigma F = ma$$

$$F_{chain} - mg = ma$$
You probably made an error in your manipulation.

3. Sep 9, 2009

### Zhalfirin88

Well I called down the positive direction so it would be

$$F_{chain} + mg = -ma$$

Correct? Isn't that what I did above?

4. Sep 9, 2009

### Fightfish

Yup, but that would yield $$F_{chain} = - (ma + mg)$$, which is not what your answer in the first post was.

5. Sep 9, 2009

### Zhalfirin88

How is it not?

6. Sep 9, 2009

### Fightfish

Your answer, $$(3 * .09kg)(9.8 m/s^2) - (3 * .09kg)(3.6 m/s^2)$$, is $$mg - ma$$.

7. Sep 9, 2009

### Zhalfirin88

Why are you adding them together when they are in opposite directions? That is why I subtracted them. I only have 1 try left or I don't get credit for this part, so I'm being a little picky, sorry about that.

8. Sep 9, 2009

### Fightfish

Well, quantitatively, if you follow my working (and even your own equation in #3), it can be shown that that should be the case. In this case, Fchain is not given by the (vector) addition of the weight of the chain and the resultant force on it. It is the (vector) addition of the weight of the chain and Fchain that yields the resultant force, and hence
$$F_{chain} + mg = -ma$$ if you take downwards as positive. (you will get a negative answer for Fchain, indicating that it is acting upwards).
(a free-body diagram will greatly aid here)
Qualitatively, the Fchain has to both overcome the weight of the chain as well as provide an upward acceleration of the chain.

9. Sep 9, 2009

### Zhalfirin88

So the equation would be:

$$F = -( (3 *.09kg)(3.6 m/s^2) + ((3 * .09kg)(9.8 m/s^2))$$

Correct answer F = -3.618 N ?

10. Sep 9, 2009

### Fightfish

Seems like there's a problem with your LaTeX code, but yup the equation in the raw code looks right.

11. Sep 9, 2009

### Zhalfirin88

It's working fine now, double check? Took out the \frac which I think was the problem.

12. Sep 9, 2009

### Fightfish

Yup, assuming you took g as 9.8.

13. Sep 10, 2009

### Zhalfirin88

Okay, new problem that uses Newton's 2nd Law.

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

A 65.0 kg girl weighs herself by standing on a scale in an elevator. What is the force exerted by the scale when the elevator is descending at a constant speed of 10 m/s?
(b)What is the force exerted by the scale if the elevator is accelerating downward with an acceleration of 2.4 m/s2?
(c)If the elevator's descending speed is measured at 10 m/s at a given point, but its speed is decreasing by 2.4 m/s2, what is the force exerted by the scale?

3. The attempt at a solution
(a) $$F = ma$$
$$F = (65.0kg)(9.8 m/s^2)$$
$$F = 637 N$$ got the right answer. However it's b and c that I'm confused about. How does the acceleration change the force?

14. Sep 10, 2009

### rl.bhat

b) F = m(g - a)

15. Sep 10, 2009

### Zhalfirin88

That's true for (b) but what about (c)? How does the velocity factor in? It gives you the deceleration but why isn't it the same as (b)?

16. Sep 10, 2009

### Fightfish

The magnitude of the velocity has no implications on the force or acceleration of the body in this scenario. In this case, it is the direction of the velocity that matters - the elevator is descending - and its speed is decreasing, which implies that the elevator is accelerating upwards.
For (b), the elevator is accelerating downwards, thus the answers for (b) and (c) would definitely differ.

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