Newton's Laws and Force Problem

AI Thread Summary
The discussion revolves around a physics problem involving a chain of three identical links, each weighing 380g, suspended and pulled upward with a force of 12N. The student successfully calculates the acceleration of the chain as 0.726 meters per second squared using Newton's second law (F=ma). However, they struggle with determining the force exerted by the top link on the middle link. Suggestions include drawing free body diagrams for each link to visualize the forces acting on them. Understanding these diagrams is essential for solving part B of the problem.
Yosty22
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Homework Statement


A student tries to raise a chain consisting of three identical links. Each link has a mass of 380g. The three-piece chain is connected to a 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 of 12N is applied to the chain by the string. Use Newton's laws to answer the following questions.
a). Find the acceleration of the chain.
b). Find the force exerted by the top link on the middle link.

Homework Equations



F=ma
F_A on B= -F_B on A

The Attempt at a Solution



I can't for the life of me figure out letter b. For a), I did:
T-mg=ma
12-(.380*3)(9.8)=(0.380*3)a
Solve for a, get:
acceleration = 0.726 meters per second squared.
This is the correct answer. However, I have no idea at all how to set up or solve part B.
 
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Yosty22 said:

Homework Statement


A student tries to raise a chain consisting of three identical links. Each link has a mass of 380g. The three-piece chain is connected to a 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 of 12N is applied to the chain by the string. Use Newton's laws to answer the following questions.
a). Find the acceleration of the chain.
b). Find the force exerted by the top link on the middle link.


Homework Equations



F=ma
F_A on B= -F_B on A

The Attempt at a Solution



I can't for the life of me figure out letter b. For a), I did:
T-mg=ma
12-(.380*3)(9.8)=(0.380*3)a
Solve for a, get:
acceleration = 0.726 meters per second squared.
This is the correct answer. However, I have no idea at all how to set up or solve part B.
Draw a Free Body Diagram for the bottom link.

Draw a Free Body Diagram for the middle link. (Maybe this one is all you need.)

Draw a Free Body Diagram for the top link.
 
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