What Force Does the Top Link Exert on the Middle Link in a Suspended Chain?

  • Thread starter Thread starter terryds
  • Start date Start date
  • Tags Tags
    Chain Force Link
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving a chain of three identical links, each with a mass of 200 g, suspended vertically and subjected to an upward force of 15.0 N. Participants are tasked with determining the force exerted by the top link on the middle link.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the application of Newton's laws, particularly the net force and acceleration of the system. There is discussion about the forces acting on the links and the relationship between the forces exerted by the links on each other.

Discussion Status

Participants have provided various calculations and clarifications regarding the forces involved. There is ongoing exploration of the relationship between the forces exerted by the links, with some participants questioning the assumptions made about the system's acceleration and the gravitational force used in calculations.

Contextual Notes

Some participants suggest using an alternative value for gravitational acceleration (g=10 m/s²) for simplification. There is also mention of the assumption that the chain is not slack, which affects the tension between the links.

terryds
Messages
392
Reaction score
13

Homework Statement



A student tries to raise a chain consisting of three identical links. Each link has a mass of 200 g. The three-piece chain is connected to light string and then suspended vertically, with the student holding the upper end of the string and pulling upward. Because of the student's pull, an upward force 15.0 N is applied to the chain by the string. Find the force exerted by the top link on the middle link.
A) 3.0 N
B) 6.0 N
C) 8.0 N
D) 10.0 N
E) None of the above

Homework Equations



ΣF = ma

The Attempt at a Solution



(The system)
ΣF = ma
15 - 3 mg = 3 ma
15 - 3 * 0.2 * 9.8 = 3 * 0.2 a
a = 15.2 m/s^2

(The top link)
ΣF = ma
15 - F = 0.2 * 15.2
F = 15 - 3.04 = 11.96 N

Is it right? I'm not sure about this
 
Physics news on Phys.org
A small clarification in -
"
The top link)
ΣF = ma
15 - F = 0.2 * 15.2
F = 15 - 3.04 = 11.96 N"
What exactly is F?
 
  • Like
Likes   Reactions: terryds
Suraj M said:
A small clarification in -
"
The top link)
ΣF = ma
15 - F = 0.2 * 15.2
F = 15 - 3.04 = 11.96 N"
What exactly is F?
The force the top link exerts on the middle link
 
terryds said:
(The system)
ΣF = ma
15 - 3 mg = 3 ma
15 - 3 * 0.2 * 9.8 = 3 * 0.2 a
a = 15.2 m/s^2

(The top link)
ΣF = ma
15 - F = 0.2 * 15.2
F = 15 - 3.04 = 11.96 N

That looks good. Note that F in your calculation represents the force that the middle link exerts on the top link. The question asks for the force that the top link exerts on the middle link.

Are they the same? Why?
 
The force applied by one part of the chain on another, at any point, is the weight of the portion of the chain below that point.
 
HallsofIvy said:
The force applied by one part of the chain on another, at any point, is the weight of the portion of the chain below that point.
...only if the chain is not accelerating (assuming "weight" refers to the force of gravity).
 
terryds said:
15 - 3 mg = 3 ma
15 - 3 * 0.2 * 9.8 = 3 * 0.2 a
I suspect that for the purposes of this question you need to use g=10m/s2.
The quickest way to the answer involves a noninertial frame.
 
haruspex said:
I suspect that for the purposes of this question you need to use g=10m/s2.
The quickest way to the answer involves a noninertial frame.

15 - 3 mg = 3 ma
15 - 3 * 0.2 * 10 = 3 * 0.2 * a
9 = 0.6 a
a = 15 m/s^2

(the top link)
ΣF = ma
15 - F - mg = 0.2 * 15
15 - 0.2 * 10 - F = 3
F = 15 - 2 -3 = 10 N

TSny said:
That looks good. Note that F in your calculation represents the force that the middle link exerts on the top link. The question asks for the force that the top link exerts on the middle link.

Are they the same? Why?

It's the same because it is not slack.. In other words, the tension must be equal..
 
TSny said:
Note that F in your calculation represents the force that the middle link exerts on the top link. The question asks for the force that the top link exerts on the middle link.

Are they the same? Why?
terryds said:
It's the same because it is not slack.. In other words, the tension must be equal..
One of Newton's laws of motion is relevant here.
 
  • #10
terryds said:
15 - 3 mg = 3 ma
15 - 3 * 0.2 * 10 = 3 * 0.2 * a
9 = 0.6 a
a = 15 m/s^2

(the top link)
ΣF = ma
15 - F - mg = 0.2 * 15
15 - 0.2 * 10 - F = 3
F = 15 - 2 -3 = 10 N
.
Yes.
The noninertial frame method uses the chain as the reference frame. The acceleration then gives rise to an inertial force that looks like extra gravity. So we have a static chain under increased gravity weighing a total of 15N. Each link therefore weighs 5N.
 
  • Like
Likes   Reactions: terryds
  • #11
TSny said:
One of Newton's laws of motion is relevant here.
Yup... The Newton 3 law (magnitude of action = reaction)
 
  • #12
haruspex said:
I suspect that for the purposes of this question you need to use g=10m/s2.
The result is independent of g.
It is independent of the three masses as well, as long as all three pieces have the same mass.

Probably the shortest way:
No matter what the actual acceleration value is, accelerating three identical links needs 3/2 the force to accelerate two links. And 3/2 of the force the question asks for are 15 N...
 
  • Like
Likes   Reactions: terryds and TSny

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
Replies
14
Views
6K
  • · Replies 15 ·
Replies
15
Views
11K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K