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

In summary, 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. The top link exerts a force of 10.0 N on the middle link.
  • #1
terryds
392
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
  • #2
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 terryds
  • #3
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
 
  • #4
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?
 
  • #5
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.
 
  • #6
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).
 
  • #7
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.
 
  • #8
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..
 
  • #9
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 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 terryds and TSny

What is top link force on chain?

Top link force on chain refers to the amount of force that is exerted on the top link of a chain when it is being pulled or loaded. This force is usually measured in pounds or Newtons and is an important factor to consider when designing or using chains in various applications.

Why is top link force on chain important?

The top link force on chain is important because it affects the overall strength and durability of the chain. If the force exceeds the maximum capacity of the chain, it can lead to deformation, breakage, or failure of the chain. It is also important to consider top link force when determining the appropriate chain for a specific task.

How is top link force on chain measured?

Top link force on chain is typically measured using a dynamometer, which is a device that can measure force and tension. The chain is attached to the dynamometer and pulled until the maximum force is reached. The reading on the dynamometer indicates the top link force on the chain.

What factors can affect top link force on chain?

There are several factors that can affect the top link force on chain, including the weight of the load, the angle of the chain, the type of chain, and the condition of the chain. These factors should be taken into consideration when determining the appropriate chain for a specific task to ensure the safety and efficiency of the chain.

How can top link force on chain be minimized?

To minimize top link force on chain, it is important to choose the correct type and size of chain for the task at hand. It is also important to properly maintain and inspect the chain to ensure it is in good condition. Additionally, using proper techniques and tools when handling and pulling the chain can help to reduce the force exerted on the top link.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
14
Views
5K
  • Introductory Physics Homework Help
Replies
15
Views
10K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
5K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top