Force of Link 3 on Link 2: 2.964 N

In summary, the force exerted by the scale on the girl in the elevator depends on the acceleration of the elevator, which is influenced by both the mass of the girl and the speed and direction of the elevator.
  • #1
Zhalfirin88
137
0

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]
 
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  • #2
[tex]\Sigma F = ma[/tex]

[tex]F_{chain} - mg = ma[/tex]
You probably made an error in your manipulation.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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?
 
  • #6
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].
 
  • #7
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
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.
 
  • #9
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.
 

What is the Force of Link 3 on Link 2?

The force of Link 3 on Link 2 refers to the amount of force that Link 3 exerts on Link 2 in a given system or scenario. It is typically measured in Newtons (N).

What factors affect the Force of Link 3 on Link 2?

The force of Link 3 on Link 2 can be affected by several factors, including the mass and acceleration of Link 3, the angle of the linkages, and any external forces acting on the system.

How is the Force of Link 3 on Link 2 calculated?

The force of Link 3 on Link 2 can be calculated using Newton's Second Law of Motion, which states that force is equal to mass multiplied by acceleration. The angle and direction of the force must also be taken into account when calculating the overall force on Link 2.

Why is the Force of Link 3 on Link 2 important in a system?

The force of Link 3 on Link 2 is important because it helps to determine the overall forces and movements within a system. Understanding this force can help with predicting and controlling the motion of the system and its individual components.

How does the Force of Link 3 on Link 2 affect the stability of a system?

The force of Link 3 on Link 2 can affect the stability of a system by influencing its equilibrium. If the force is too great, it can cause the system to become unbalanced and potentially lead to failure or malfunction. Properly controlling and balancing this force is crucial for maintaining stability in the system.

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