Finding Tension in a Chain: Understanding Torque and Force Components

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Homework Help Overview

The discussion revolves around finding the tension in a chain and understanding the concepts of torque and force components in a physics problem involving a rod and a chain. Participants explore the relationships between forces, angles, and torques in the context of static equilibrium.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of torque equations and the need to define unknown variables. There are questions about the angle at which forces act and the implications of different configurations of the chain and rod. Some participants suggest drawing diagrams to clarify the forces involved.

Discussion Status

The discussion is active, with participants providing guidance on how to approach the problem analytically. There is a focus on equating torques and understanding the roles of different forces. Some participants express uncertainty about specific angles and their effects on torque calculations.

Contextual Notes

Participants note the importance of understanding the center of mass of the rod and how it relates to the torque due to weight. There is also mention of the need to clarify the angle of the chain and its effect on the calculations.

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Homework Statement



J5T6pCf.png

Homework Equations

The Attempt at a Solution


seems like Torque = r f sin theta should be used here but we don't know r.

Am I suppose to find the x and y components of the tension of the chain?

off topic : if the chain was moved so it would just be above the end of the rod, just like this :
nFy8K4l.png
, would it put less tension on the chain?
 
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For variables you don't know, adopt a symbol and go to work. Good chance the thing will drop out towards the end.
 
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goonking said:
seems like Torque = r f sin theta should be used here but we don't know r.

You do not always need to know all of the quantities. Why don't you start by simply assuming that the length of the rod is R and do things analytically from there. A good practice is to never try to insert numbers until you have arrived at your final expression.
 
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No the chain is in the proper angle to help you calculate the torque of the tension of chain easily ##T_1=31.2r## where r is the length of rod. This torque will be countered by the torque of the weight. Can you find what is the angle ##\phi_1## that the weight makes with the line of rod and then find the torque of weight? You then can see that r is simplified when equating the two torques.
 
Problem solving systematics urges you to list the relevant equations and the given/known data.
You do know them already and you do use them already, but still it's a good way to get a grasp on an exercise.

In this case you want ##\Sigma\;\tau = 0## and indeed need ##\tau = \vec r\times\vec F = r F \sin\theta## but this ##\theta## is not the ##\theta## in the drawing that comes with your exercise.

Oh, am I a slow typist. And I need to do some paid work too, so I leave you to the other good helpers...
 
Delta² said:
No the chain is in the proper angle to help you calculate the torque of the tension of chain easily ##T_1=31.2r## where r is the length of rod. This torque will be countered by the torque of the weight. Can you find what is the angle ##\phi_1## that the weight makes with the line of rod and then find the torque of weight? You then can see that r is simplified when equating the two torques.
i'm not sure where you mean ϕ1. by weight, do you mean the right end of the rod?
 
goonking said:
i'm not sure where you mean ϕ1. by weight, do you mean the right end of the rod?

No i mean the force of weight due to gravity... Can you draw a diagram with all the forces acting on the rod?
 
Delta² said:
No i mean the force of weight due to gravity... Can you draw a diagram with all the forces acting on the rod?
should be 78 degrees
 
Ok correct, now what is the torque of weight?
 
  • #10
Delta² said:
Ok correct, now what is the torque of weight?
r f sin 78

f = mg

?
 
  • #11
Not exactly. The force of weight is considered to be applied at the c.o.m of the rod. Where is the c.o.m of rod?
 
  • #12
Delta² said:
Not exactly. The force of weight is considered to be applied at the c.o.m of the rod. Where is the c.o.m of rod?
at the middle of the rod
 
  • #13
ok so you have the weight acting on the middle of the rod, at an agle of 78, so what is its torque?
 
  • #14
Delta² said:
ok so you have the weight acting on the middle of the rod, at an agle of 78, so what is its torque?
r/2 x mg x sin 78

?
 
  • #16
Delta² said:
yes correct.
do we need the torque of the chain? or is that given by the tension in the chain?
 
  • #17
yes you need the torque of the chain. I already said in post #4 what it is. Equate the two torques and solve for m. What do you get?
 
  • #18
Delta² said:
yes you need the torque of the chain. I already said in post #4 what it is. Equate the two torques and solve for m. What do you get?
torque of chain = r 31.2

r 31.2N = r mg sin 78

m = 3.544kg

?
 
  • #19
Didnt we say the torque of weight is (r/2)mgsin(78)
 
  • #20
Delta² said:
Didnt we say the torque of weight is (r/2)mgsin(78)
whoops, yes
 
  • #21
Delta² said:
Didnt we say the torque of weight is (r/2)mgsin(78)

r 31.2N = r/2 mg sin 78

m = 6.5 kg

?
 
  • #22
yes correct. what value did u use for g just asking.
 
  • #23
Delta² said:
yes correct. what value did u use for g just asking.
9.8
 
  • #24
still having a bit of trouble understanding why torque = 31.2 r

isn't the chain at an 78 degree angle too?
 
  • #25
goonking said:
isn't the chain at an 78 degree angle too?

No. The force from the chain is acting perpendicular to the rod and therefore has the full length of the rod as the moment arm.

Edit: It would have a 78 degree angle if it was positioned as in your drawing, but it is not.
 
  • #26
Orodruin said:
No. The force from the chain is acting perpendicular to the rod and therefore has the full length of the rod as the moment arm.

Edit: It would have a 78 degree angle if it was positioned as in your drawing, but it is not.
i see, thank you.
 

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