Force of Link 3 on Link 2: 2.964 N

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Homework Help Overview

The discussion revolves around a physics problem involving a chain of links being lifted with a constant acceleration. The original poster seeks to determine the force exerted by one link on another within the chain, specifically focusing on the forces acting on link 2 due to link 3.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law and the correct setup of force equations. There are attempts to clarify the direction of forces and the resultant equations based on the chosen coordinate system.

Discussion Status

Participants are actively engaging in clarifying the equations and addressing potential errors in manipulation. There is a focus on understanding the relationship between the forces acting on the links and the acceleration of the chain. Some guidance has been offered regarding the interpretation of forces and the use of free-body diagrams.

Contextual Notes

There is a concern about the accuracy of calculations and the implications of directionality in force equations. The original poster expresses urgency due to limited attempts remaining for the assignment.

Zhalfirin88
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Homework Statement


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

Homework Equations


[tex]\Sigma F = ma[/tex]

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.

[tex](3 * .09kg)(9.8 m/s^2) - (3 * .09kg)(3.6 m/s^2)[/tex]
 
Last edited:
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[tex]\Sigma F = ma[/tex]

[tex]F_{chain} - mg = ma[/tex]
You probably made an error in your manipulation.
 
Fightfish said:
[tex]\Sigma F = ma[/tex]

[tex]F_{chain} - mg = ma[/tex]
You probably made an error in your manipulation.

Well I called down the positive direction so it would be

[tex]F_{chain} + mg = -ma[/tex]

Correct? Isn't that what I did above?
 
Zhalfirin88 said:
Well I called down the positive direction so it would be

[tex]F_{chain} + mg = -ma[/tex]

Correct? Isn't that what I did above?
Yup, but that would yield [tex]F_{chain} = - (ma + mg)[/tex], which is not what your answer in the first post was.
 
Fightfish said:
Yup, but that would yield [tex]F_{chain} = - (ma + mg)[/tex], which is not what your answer in the first post was.

How is it not?
 
Zhalfirin88 said:
How is it not?
Your answer, [tex](3 * .09kg)(9.8 m/s^2) - (3 * .09kg)(3.6 m/s^2)[/tex], is [tex]mg - ma[/tex].
 
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.
 
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
[tex]F_{chain} + mg = -ma[/tex] 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.
 
So the equation would be:

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

Correct answer F = -3.618 N ?
 
  • #10
Seems like there's a problem with your LaTeX code, but yup the equation in the raw code looks right.
 
  • #11
Fightfish said:
Seems like there's a problem with your LaTeX code, but yup the equation in the raw code looks right.

It's working fine now, double check? Took out the \frac which I think was the problem.
 
  • #12
Yup, assuming you took g as 9.8.
 
  • #13
Okay, new problem that uses Newton's 2nd Law.

1. Homework Statement

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?

The Attempt at a Solution


(a) [tex]F = ma[/tex]
[tex]F = (65.0kg)(9.8 m/s^2)[/tex]
[tex]F = 637 N[/tex] got the right answer. However it's b and c that I'm confused about. How does the acceleration change the force?
 
  • #14
b) F = m(g - a)
 
  • #15
rl.bhat said:
b) F = m(g - a)

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
Zhalfirin88 said:
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)?
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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